Find closed form for quadruple integral

Question

I am trying to find a closed form of the following integral $$ \int _0^{\infty }\int _0^x\int _0^y\int _0^z \exp \left( -\frac{a x^2}{2}-\frac{b y^2}{2}-\frac{c z^2}{2}-\frac{d w^2}{2} \right) \,\mathrm{d}w\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x $$ where $a,b,c,d>0$ are some constants.

My idea is to change variable to to polar system by letting $$ x=\frac{r \cos (\alpha ) \cos (\beta ) \cos (\theta )}{\sqrt{a}}, \quad y=\frac{r \cos (\alpha ) \cos (\beta ) \sin (\theta )}{\sqrt{a}} $$ $$ z=\frac{r \sin (\alpha ) \cos (\beta )}{\sqrt{c}}, \quad w=\frac{r \sin (\beta )}{\sqrt{d}} $$

This reduces the original integral into $$ \int_0^{\tan ^{-1}\left(\sqrt{\frac{b}{a}}\right)} \frac{\sin (\theta ) \tan ^{-1}\left(\sin (\theta ) \sqrt{\frac{d}{b+c \sin ^2(\theta )}}\right)}{\sqrt{a b d \left(b+c \sin ^2(\theta )\right)}} \, d\theta $$ But then I get stuck here.

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PS: I am interested in this because I found that $$ \int _0^{\infty }\int _0^x \exp \left(-\frac{a x^2}{2}-\frac{b y^2}{2}\right)\,\mathrm{d}y\,\mathrm{d}x = \frac{\tan ^{-1}\left(\sqrt{\frac{b}{a}}\right)}{\sqrt{a b}} $$ and $$ \int _0^{\infty }\int _0^x\int _0^y \exp \left( -\frac{a x^2}{2}-\frac{b y^2}{2}-\frac{c z^2}{2} \right) \,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x = \frac{\sqrt{\pi/2 } }{\sqrt{a b c}} \left(\tan ^{-1}\left(\sqrt{\frac{c}{b}}\right)-\tan ^{-1}\left(\sqrt{\frac{a c}{b (a+b+c)}}\right)\right) $$ So I am trying to generalize this. Maybe this is already known?

Answer 1

Define: \begin{eqnarray} I(a,b,c,d):=\int_0^{\tan ^{-1}\left(\sqrt{\frac{b}{a}}\right)} \frac{\sin (\theta ) \tan ^{-1}\left(\sin (\theta ) \sqrt{\frac{d}{b+c \sin ^2(\theta )}}\right)}{\sqrt{a b d \left(b+c \sin ^2(\theta )\right)}} \, d\theta \quad (i) \end{eqnarray} Then also define:

\begin{eqnarray} {\mathfrak F}^{(A,B)}_{a,b} &:=& \int\limits_A^B \frac{\log(z+a)}{z+b} dz\\ &=& F[B,a,b] - F[A,a,b] + 1_{t^* \in (0,1)} \left( -F[A+(t^*+\epsilon)(B-A),a,b] + F[A+(t^*-\epsilon)(B-A),a,b] \right) \quad (ii) \end{eqnarray} where \begin{eqnarray} t^*:=-\frac{Im[(A+b)(b^*-a^*)]}{Im[(B-A)(b^*-a^*)]} \end{eqnarray} and \begin{equation} F[z,a,b] := \log(z+a) \log\left( \frac{z+b}{b-a}\right) + Li_2\left( \frac{z+a}{a-b}\right) \end{equation} for $a$,$b$,$A$,$B$ being complex.

Then we have: \begin{eqnarray} I(a,b,c,d)&=& \frac{1}{\sqrt{a}} \int\limits_0^{\sqrt{\frac{d}{a+b+c}}} \frac{u \tan ^{-1}(u)}{\left(d-c u^2\right) \sqrt{d-u^2 (b+c)}} du\\ &=& \frac{\sqrt{d}}{\sqrt{a} (b+c)}\int\limits_0^{\sin ^{-1}\left(\sqrt{\frac{b+c}{a+b+c}}\right)} \frac{\sin (\phi ) \tan ^{-1}\left(\sin (\phi ) \sqrt{\frac{d}{b+c}}\right)}{d-\frac{c d \sin ^2(\phi )}{b+c}} d\phi\\ &=&-\frac{2 i}{\sqrt{a} \sqrt{d}} \int\limits_0^{\frac{\sqrt{\frac{b+c}{a+b+c}}}{\sqrt{\frac{a}{a+b+c}}+1}} \frac{t}{b \left(t^2+1\right)^2+c \left(t^2-1\right)^2} \log \left(\frac{2 i t \sqrt{\frac{d}{b+c}}+t^2+1}{-2 i t \sqrt{\frac{d}{b+c}}+t^2+1}\right) dt\\ &=&\frac{1}{4} \frac{1}{\sqrt{a b c d}} \sum\limits_{\xi=1}^4 \sum\limits_{\eta=1}^4 (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\left\lfloor \frac{\xi -1}{2}\right\rfloor } \int\limits_0^{\frac{\sqrt{\frac{b+c}{a+b+c}}}{\sqrt{\frac{a}{a+b+c}}+1}} \frac{\log \left(i (-1)^{\left\lfloor \frac{\xi -1}{2}\right\rfloor } \sqrt{\frac{d}{b+c}}+i (-1)^{\xi -1} \sqrt{\frac{b+c+d}{b+c}}+t\right)}{t-i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\eta +1} e^{i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor } \tan ^{-1}\left(\frac{\sqrt{c}}{\sqrt{b}}\right)}} dt \\ &=&\frac{1}{4} \frac{1}{\sqrt{a b c d}} \sum\limits_{\xi=1}^4 \sum\limits_{\eta=1}^4 (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\left\lfloor \frac{\xi -1}{2}\right\rfloor } % {\mathfrak F}^{(0,\frac{\sqrt{\frac{b+c}{a+b+c}}}{\sqrt{\frac{a}{a+b+c}}+1})}_{i (-1)^{\left\lfloor \frac{\xi -1}{2}\right\rfloor } \sqrt{\frac{d}{b+c}}+i (-1)^{\xi -1} \sqrt{\frac{b+c+d}{b+c}},-i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\eta +1} e^{i (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor } \tan ^{-1}\left(\frac{\sqrt{c}}{\sqrt{b}}\right)}} \end{eqnarray}

In the top line we substituted for $u=\sin(\theta) \sqrt{d/(b+c \sin(\theta)^2)}$. In the second line we substituted $u = \sqrt{d/(c+b)} \sin(\phi)$. In the third line we substituted $t=\tan(\phi/2)$. In the forth line we used partial fraction decomposition and properties of the logarithm. Finally in the fifth line we used the anti-derivative defined in $(ii)$.

Clear[F]; Clear[FF];
F[z_, a_, b_] := 
  Log[a + z] Log[(b + z)/(-a + b)] + PolyLog[2, (a + z)/(a - b)];
FF[A_, B_, a_, b_] := 
  Module[{result, ts, zs, zsp, zsm, eps = 10^(-15)},
   (*This is Integrate[Log[z+a]/(z+b),{z,A,B}] where all a,b,A, 
   and B are complex. *)
   result = F[B, a, b] - F[A, a, b];


   ts = - (Im[(A + b) (Conjugate[b] - Conjugate[a])]/
     Im[(B - A) (Conjugate[b] - Conjugate[a])]);
   If[0 <= ts <= 1,
    zsp = A + (ts + eps) (B - A);
    zsm = A + (ts - eps) (B - A);
    result += -F[zsp, a, b] + F[zsm, a, b];
    ];

   result
   ];

{a, b, c, d} = RandomReal[{0, 3}, 4, WorkingPrecision -> 50];
NIntegrate[
 Exp[-a/2 x^2 - b/2 y^2 - c/2 z^2 - d/2 w^2], {x, 0, Infinity}, {y, 0,
   x}, {z, 0, y}, {w, 0, z}]
NIntegrate[
 Sin[th]/Sqrt[a b d (b + c Sin[th]^2)] ArcTan[
   Sin[th] Sqrt[d/(b + c Sin[th]^2)]], {th, 0, ArcTan[Sqrt[b/a]]}]
  1/Sqrt[a ] NIntegrate[
  u ArcTan[u] 1/((d - c u^2) Sqrt[d - (c + b) u^2]), {u, 0, Sqrt[ 
   d/ (a + b + c)]}]
 Sqrt[d]/(Sqrt[a ] (b + c))
  NIntegrate[
  Sin[phi] ArcTan[
    Sqrt[d/(c + b)] Sin[phi]] 1/(d - c (d/(c + b) Sin[phi]^2)) , {phi,
    0, ArcSin[ Sqrt[( c + b)/ (a + b + c)]]}]
- I 2/(Sqrt[a ] Sqrt[d])
  NIntegrate[
   t /(c (-1 + t^2)^2 + b (1 + t^2)^2) Log[(
    1 + t^2 + 2 I Sqrt[d/(b + c)] t)/(
    1 + t^2 - 2 I Sqrt[d/(b + c)] t)], {t, 0, Sqrt[(b + c)/(
   a + b + c)]/(1 + Sqrt[a/(a + b + c)])}]
- I 2/(Sqrt[a ] Sqrt[d])
  NIntegrate[
   t /(c (-1 + t^2)^2 + 
     b (1 + t^2)^2) Log[((1/
        2 (2 I Sqrt[d/(b + c)] - Sqrt[-4 - (4 d)/(b + c)]) + 
       t) (1/2 (2 I Sqrt[d/(b + c)] + Sqrt[-4 - (4 d)/(b + c)]) + 
       t))/((1/2 (-2 I Sqrt[d/(b + c)] - Sqrt[-4 - (4 d)/(b + c)]) + 
       t) (1/2 (-2 I Sqrt[d/(b + c)] + Sqrt[-4 - (4 d)/(b + c)]) + 
       t))], {t, 0, Sqrt[(b + c)/(a + b + c)]/(
   1 + Sqrt[a/(a + b + c)])}]
 1/Sqrt[a b c d]  1/4 NIntegrate[
  Sum[(-1)^(Floor[(eta - 1)/2]) (-1)^
    Floor[(xi - 1)/2] Log[
     t + (-1)^Floor[(xi - 1)/2] I Sqrt[d/(b + c)] + (-1)^(xi - 1)
        I Sqrt[( b + c + d)/(b + c)]]/(
    t - (-1)^(1 + eta + 
        Floor[(eta - 1)/2]) I Exp[(-1)^(Floor[(eta - 1)/2]) I ArcTan[
         Sqrt[c]/Sqrt[b]]]), {xi, 1, 4}, {eta, 1, 4}], {t, 0, Sqrt[(
   b + c)/(a + b + c)]/(1 + Sqrt[a/(a + b + c)])}]
 1/Sqrt[a b c d]  1/4 Sum[(-1)^(Floor[(eta - 1)/2]) (-1)^
   Floor[(xi - 1)/2] FF[0, Sqrt[(b + c)/(a + b + c)]/(
    1 + Sqrt[a/(
     a + b + c)]), (-1)^Floor[(xi - 1)/2] I Sqrt[d/(b + c)] + (-1)^(
      xi - 1) I Sqrt[( b + c + d)/(
      b + c)], -(-1)^(1 + eta + 
        Floor[(eta - 1)/2]) I Exp[(-1)^(Floor[(eta - 1)/2]) I ArcTan[
        Sqrt[c]/Sqrt[b]]]], {xi, 1, 4}, {eta, 1, 4}]

Update: As a sanity check look at the case $a=b=c=d=1$. Then define: \begin{eqnarray} M1&:=&\left( \begin{array}{cccc} -1+\sqrt{3} & \sqrt{2} & \sqrt{2} & -1+\sqrt{3} \\ 1 & \sqrt{2-\sqrt{3}} & \sqrt{2-\sqrt{3}} & 1 \\ \sqrt{2-\sqrt{3}} & 1 & 1 & \sqrt{2-\sqrt{3}} \\ \sqrt{2} & -1+\sqrt{3} & -1+\sqrt{3} & \sqrt{2} \\ \end{array} \right)\\ M2&:=&\left( \begin{array}{cccc} \frac{1}{\sqrt{2}} & \frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{2} \left(-1+\sqrt{3}\right) & \frac{1}{2} \left(-1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} \\ \frac{1}{2} \left(-1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{2} \left(-1+\sqrt{3}\right) \\ \frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{2} \left(1+\sqrt{3}\right) \\ \end{array} \right)\\ A1&:=&\left( \begin{array}{cccc} -\frac{\pi }{6} & \frac{\pi }{12} & -\frac{\pi }{4} & 0 \\ \frac{5 \pi }{6} & \frac{\pi }{12} & \frac{5 \pi }{12} & -\frac{\pi }{3} \\ -\frac{5 \pi }{12} & \frac{\pi }{3} & -\frac{5 \pi }{6} & -\frac{\pi }{12} \\ \frac{\pi }{4} & 0 & \frac{\pi }{6} & -\frac{\pi }{12} \\ \end{array} \right)\\ A2&:=&\left( \begin{array}{cccc} -\frac{\pi }{12} & \frac{\pi }{6} & -\frac{\pi }{6} & \frac{\pi }{12} \\ \frac{7 \pi }{12} & -\frac{\pi }{6} & \frac{\pi }{6} & -\frac{7 \pi }{12} \\ -\frac{\pi }{6} & \frac{7 \pi }{12} & -\frac{7 \pi }{12} & \frac{\pi }{6} \\ \frac{\pi }{6} & -\frac{\pi }{12} & \frac{\pi }{12} & -\frac{\pi }{6} \\ \end{array} \right) \end{eqnarray} and we have \begin{eqnarray} I(1,1,1,1)=\frac{1}{4} \sum\limits_{\xi=1}^4 \sum\limits_{\eta=1}^4 (-1)^{\left\lfloor \frac{\eta -1}{2}\right\rfloor +\left\lfloor \frac{\xi -1}{2}\right\rfloor } \left( Li_2(M1_{\xi,\eta}\exp(\imath A1_{\xi,\eta}))- Li_2(M2_{\xi,\eta}\exp(\imath A2_{\xi,\eta})) \right) \end{eqnarray} We have checked numerically that this quantity above coincides with $\pi^2/96$ to one hundred digits. It would be interesting to prove this analytically.

Answer 2

This integral is up to normalization constant integral of multivariate gaussian distribution. Due too lack of cross terms we get that there are 4 independent zero mean gaussian variables involved.

We can begin by rewriting this a probability of an event.

Let $X,Y,Z,W$ independent normally distributed with zero mean and possibly different variances.

Then integral reduces to: $$ \mathbb{P} (0 < W < Z < Y < X) $$

Which in turn is the same as:

$$ \mathbb{P}(W > 0 \land Z-W >0 \land Y-Z >0 \land X-Y >0) $$

and one can observe that $W,Z-W,Y-Z,X-Y$ are correlated jointly normal variables. Probability that all components of a jointly normal vector are positive is called orthant probability and in general doesn't have closed form expression. I suppose this is quite special case with covariance matrix almost diagonal so maybe there are some articles about how to approach this special case.

For the case with 3 or less variables formulas are known (cf this question for example) and I guess they would coincide with what you've found.

Answer 3

Here we give an answer using a different method. Assume that $a\ge0$, $b\ge 0$, $c\ge 0$ and $d\ge 0$. Define: \begin{equation} I(a,b,c,d):=\int _0^{\infty }\int _0^x\int _0^y\int _0^z \exp \left( -\frac{a x^2}{2}-\frac{b y^2}{2}-\frac{c z^2}{2}-\frac{d w^2}{2} \right) \,\mathrm{d}w\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x \end{equation} Then we have: \begin{eqnarray} &&I(a,b,c,d)=\\ &&\sqrt{\frac{\pi}{2 c d}} \int\limits_0^\infty \int\limits_0^x \int\limits_0^{\sqrt{c} y}\exp(-\frac{a x^2}{2} - \frac{b y^2}{2}-\frac{ z^2}{2})\cdot \text{erf}(\sqrt{\frac{d}{2 c}} z)\; dz \; dy \;dx=\\ &&\frac{2 \pi}{\sqrt{c d}} \int\limits_0^\infty \int\limits_0^x \exp(-\frac{a x^2}{2}-\frac{b y^2}{2}) \left(\frac{1}{2\pi} \arctan[\sqrt{\frac{d}{c}}] - T(\sqrt{c} y,\sqrt{\frac{d}{c}}\right)\; dy \; dx=\\ && \frac{2 \pi^2}{\sqrt{a b c d}}\int\limits_0^\infty \frac{\exp(-\frac{y^2}{2})}{\sqrt{2\pi}} \text{erfc}(\sqrt{\frac{a}{2 b}y})\left(\frac{1}{2\pi} \arctan[\sqrt{\frac{d}{c}}] - T(\sqrt{\frac{c}{b}} y,\sqrt{\frac{d}{c}}\right) \; dy=\\ && \frac{2 \pi^2}{\sqrt{a b c d}} \left( \frac{\left(\pi -2 \tan ^{-1}\left(\sqrt{\frac{a}{b}}\right)\right) \tan ^{-1}\left(\sqrt{\frac{d}{c}}\right)-\pi \sin ^{-1}\left(\frac{\sqrt{b d}}{\sqrt{(b+c) (c+d)}}\right)}{4 \pi ^2}+ \int\limits_0^\infty \frac{\exp(-\frac{y^2}{2})}{\sqrt{2 \pi}} \text{erf}(\sqrt{\frac{a}{2 b}} y) T(\sqrt{\frac{c}{b}} y,\sqrt{\frac{d}{c}} )\; dy \right)=\\ && \frac{2 \pi^2}{\sqrt{a b c d}} \left( \frac{\tan ^{-1}\left(\sqrt{\frac{d}{c}}\right)-\tan ^{-1}\left(\frac{\sqrt{b d}}{\sqrt{c (b+c+d)}}\right)}{4 \pi } +\right.\\ &&\left. \frac{1}{8 \pi^2} \sum\limits_{i=1}^4 \sum\limits_{j=1}^4 (-1)^{\left\lfloor \frac{i-1}{2}\right\rfloor +j} {\mathfrak F}^{(1,\sqrt{\frac{a+b+c}{b+c}}-\sqrt{\frac{a}{b+c}})}_{i \left((-1)^{\left\lfloor \frac{j-1}{2}\right\rfloor } \sqrt{\frac{b+c+d}{b+c}}+(-1)^j \sqrt{\frac{d}{b+c}}\right),(-1)^{i+1} \sqrt{\frac{c}{b+c}}+i \sqrt{\frac{b}{b+c}} (-1)^{\left\lceil \frac{i-1}{2}\right\rceil +1}} \right) \end{eqnarray} In the first line we integrated over $w$ using the definition of the error function. In the second line we integrated over $z$ using the definition of the Owen's T function. In the third line we swapped the order of integration and integrated over $x$ using the definition of the complementary error function. In the forth line we split the integral into doable integrals and a more complicated one and finally in the fifth line we evaluated the remaining integral using An integral involving a Gaussian, error functions and the Owen's T function. .

Clear[F]; Clear[FF];
F[z_, a_, b_] := 
  Log[a + z] Log[(b + z)/(-a + b)] + PolyLog[2, (a + z)/(a - b)];
FF[A_, B_, a_, b_] := 
  Module[{result, ts, zs, zsp, zsm, eps = 10^(-50)},
   (*This is Integrate[Log[z+a]/(z+b),{z,A,B}] where all a,b,A, 
   and B are complex. *)
   result = F[B, a, b] - F[A, a, b];


   ts = - (Im[(A + b) (Conjugate[b] - Conjugate[a])]/
     Im[(B - A) (Conjugate[b] - Conjugate[a])]);
   If[0 <= ts <= 1,
    zsp = A + (ts + eps) (B - A);
    zsm = A + (ts - eps) (B - A);
    result += -F[zsp, a, b] + F[zsm, a, b];
    ];

   result
   ];
J[a_, b_, c_] := 
  1/ Pi^2 (ArcTan[Sqrt[2] a]/2 ArcTan[ c] + 
     1/8  Sum[
       FF[1, ( Sqrt[1 + 2 a^2 + b^2] - Sqrt[2] a)/Sqrt[
         1 + b^2], ((-1)^j I b c + (-1)^Floor[(j - 1)/2] I Sqrt[
           1 + b^2 + b^2 c^2])/Sqrt[
         1 + b^2], -(((-1)^Ceiling[(i - 1)/2] I + (-1)^i b)/Sqrt[
          1 + b^2])] (-1)^(j + Floor[(i - 1)/2]), {i, 1, 4}, {j, 1, 
        4}] );

{a, b, c, d} = RandomReal[{0, 10}, 4, WorkingPrecision -> 50];
NIntegrate[
 Exp[-a/2 x^2 - b/2 y^2 - c/2 z^2 - d/2 w^2], {x, 0, Infinity}, {y, 0,
   x}, {z, 0, y}, {w, 0, z}]
Sqrt[\[Pi]/2]/(Sqrt[c] Sqrt[d])
  NIntegrate[
  Exp[-a/2 x^2 - b/2 y^2 - 1/2 z^2] Erf[Sqrt[d/(2 c)] z], {x, 0, 
   Infinity}, {y, 0, x}, {z, 0, Sqrt[c] y}]
 (2 \[Pi])/Sqrt[c d]
  NIntegrate[
  Exp[-a/2 x^2 - b/2 y^2] (ArcTan[Sqrt[d/c]]/(2 \[Pi]) - 
     OwenT[Sqrt[c] y, Sqrt[d/c]]), {x, 0, Infinity}, {y, 0, x}]
(2  \[Pi]^2)/Sqrt[a b c d]
  NIntegrate[
  Exp[ -1/2 y^2]/Sqrt[2 Pi]
    Erfc[Sqrt[a/(2 b)] y] (ArcTan[Sqrt[d/c]]/(2 \[Pi]) - 
     OwenT[Sqrt[c/b] y, Sqrt[d/c]]), {y, 0, Infinity}]
(2  \[Pi]^2)/Sqrt[
 a b c d] ((-\[Pi] ArcSin[Sqrt[b d]/
      Sqrt[(b + c) (c + d)]] + (\[Pi] - 2 ArcTan[Sqrt[a/b]]) ArcTan[
      Sqrt[d/c]])/(4 \[Pi]^2) + 
   NIntegrate[
    Exp[ -1/2 y^2]/Sqrt[2 Pi]
      Erf[Sqrt[a/(2 b)] y] OwenT[Sqrt[c/b] y, Sqrt[d/c]], {y, 0, 
     Infinity}])
(2  \[Pi]^2)/Sqrt[
 a b c d] ((-ArcTan[Sqrt[b d]/ Sqrt[c (b + c + d)]] + 
    ArcTan[Sqrt[d/c]])/(4 \[Pi]) +  
   1/( 8 Pi^2)
     Sum[FF[1, Sqrt[(a + b + c)/(b + c)] - Sqrt[a/(b + c)], 
       I  ((-1)^j Sqrt[d/(b + c)] + (-1)^Floor[1/2 (-1 + j)] Sqrt[(
           b + c + d)/(b + c)]), (-1)^(i + 1) Sqrt[c/(b + c)] + 
        Sqrt[b/(b + c)] I (-1)^(1 + Ceiling[1/2 (-1 + i)])] (-1)^(
      j + Floor[(i - 1)/2]), {i, 1, 4}, {j, 1, 4}] )

Answer 4

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Define the function $\mathcal{I}:\mathbb{R}_{>0}^{4}\rightarrow\mathbb{R}$ via the quadruple integral

$$\mathcal{I}{\left(a,b,c,d\right)}:=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\int_{0}^{z}\mathrm{d}w\,\exp{\left(-\frac{ax^{2}}{2}-\frac{by^{2}}{2}-\frac{cz^{2}}{2}-\frac{dw^{2}}{2}\right)}.\tag{1}$$

If we look at how this integral transforms under the substitution $\left(x,y,z,w\right)\mapsto\left(px,py,pz,pw\right)$ for some fixed but arbitrary positive real $p$, we obtain the following scaling relation:

$$\forall\left(a,b,c,d,p\right)\in\mathbb{R}_{>0}^{5}:\mathcal{I}{\left(a,b,c,d\right)}=p^{4}\mathcal{I}{\left(ap^{2},bp^{2},cp^{2},dp^{2}\right)}.\tag{2}$$

As such, in our general evaluation of $\mathcal{I}{\left(a,b,c,d\right)}$ it will suffice to consider the $a=1$ case.

---

To begin with, we derive some quick integration formulas that will be helpful below.

For any $p\in\mathbb{R}_{>0}$,

$$\begin{align} \int_{0}^{\infty}\mathrm{d}x\,x^{3}\exp{\left(-\frac{px^{2}}{2}\right)} &=\int_{0}^{\infty}\mathrm{d}y\,\frac{y}{2}\exp{\left(-\frac{py}{2}\right)};~~~\small{\left[x=\sqrt{y}\right]}\\ &=\int_{0}^{\infty}\mathrm{d}z\,\frac{2}{p}\cdot\frac{z}{p}\exp{\left(-z\right)};~~~\small{\left[y=\frac{2z}{p}\right]}\\ &=\frac{2}{p^{2}}\int_{0}^{\infty}\mathrm{d}z\,z\exp{\left(-z\right)}\\ &=\frac{2}{p^{2}}.\tag{3a}\\ \end{align}$$

Next, given $p\in\mathbb{R}_{>0}$ and setting $q:=\sqrt{p}$,

$$\begin{align} \int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left(1+pt^{2}\right)^{2}} &=\int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left(1+q^{2}t^{2}\right)^{2}}\\ &=\frac{1}{q^{3}}\int_{0}^{q}\mathrm{d}x\,\frac{2x^{2}}{\left(1+x^{2}\right)^{2}};~~~\small{\left[qt=x\right]}\\ &=\frac{1}{q^{3}}\int_{0}^{q}\mathrm{d}x\,\frac{\mathrm{d}}{\mathrm{d}x}\left[\arctan{\left(x\right)}-\frac{x}{1+x^{2}}\right]\\ &=\frac{1}{q^{3}}\left[\arctan{\left(q\right)}-\frac{q}{1+q^{2}}\right]\\ &=\left[\frac{\arctan{\left(q\right)}}{q^{3}}-\frac{1}{q^{2}\left(1+q^{2}\right)}\right]\\ &=\left[\frac{\arctan{\left(\sqrt{p}\right)}}{p\sqrt{p}}-\frac{1}{p\left(1+p\right)}\right].\tag{3b}\\ \end{align}$$

Defining the function $\mathcal{J}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ by

$$\mathcal{J}{\left(p,q,x\right)}:=\int_{0}^{x}\mathrm{d}y\,\left[\frac{\arctan{\left(\sqrt{p+qy^{2}}\right)}}{\left(p+qy^{2}\right)^{3/2}}-\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\right],\tag{3c}$$

we then find that for any $\left(p,q,x\right)\in\mathbb{R}_{>0}^{3}$,

$$\begin{align} \mathcal{J}{\left(p,q,x\right)} &=\int_{0}^{x}\mathrm{d}y\,\left[\frac{\arctan{\left(\sqrt{p+qy^{2}}\right)}}{\left(p+qy^{2}\right)^{3/2}}-\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\right]\\ &=\int_{0}^{x}\mathrm{d}y\,\frac{\arctan{\left(\sqrt{p+qy^{2}}\right)}}{\left(p+qy^{2}\right)^{3/2}}-\int_{0}^{x}\mathrm{d}y\,\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\\ &=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\int_{0}^{x}\mathrm{d}y\,\frac{qy^{2}}{p\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)};~~~\small{I.B.P.}\\ &~~~~~-\int_{0}^{x}\mathrm{d}y\,\frac{1}{\left(p+qy^{2}\right)\left(1+p+qy^{2}\right)}\\ &=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\int_{0}^{x}\mathrm{d}y\,\frac{1}{p\left(1+p+qy^{2}\right)}\\ &=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\frac{1}{p\sqrt{q\left(1+p\right)}}\int_{0}^{x\sqrt{\frac{q}{1+p}}}\mathrm{d}t\,\frac{1}{\left(1+t^{2}\right)};~~~\small{\left[y=t\sqrt{\frac{1+p}{q}}\right]}\\ &=\frac{x\arctan{\left(\sqrt{p+qx^{2}}\right)}}{p\sqrt{p+qx^{2}}}-\frac{\arctan{\left(\frac{x\sqrt{q}}{\sqrt{1+p}}\right)}}{p\sqrt{q\left(1+p\right)}}.\tag{3d}\\ \end{align}$$

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Suppose $\left(b,c,d\right)\in\mathbb{R}_{>0}^{3}$.

We begin by reducing $\mathcal{I}$ to a single-variable integral in the following way:

$$\begin{align} \mathcal{I}{\left(1,b,c,d\right)} &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\int_{0}^{z}\mathrm{d}w\,\exp{\left(-\frac{x^{2}}{2}-\frac{by^{2}}{2}-\frac{cz^{2}}{2}-\frac{dw^{2}}{2}\right)}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,x^{3}\\ &~~~~~\times\exp{\left(-\frac{x^{2}}{2}-\frac{bx^{2}t^{2}}{2}-\frac{cx^{2}u^{2}}{2}-\frac{dx^{2}v^{2}}{2}\right)};~~~\small{\left[\left(y,z,w\right)=\left(xt,xu,xv\right)\right]}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,x^{3}e^{-\frac{\left(1+bt^{2}+cu^{2}+dv^{2}\right)x^{2}}{2}}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\int_{0}^{\infty}\mathrm{d}x\,x^{3}e^{-\frac{\left(1+bt^{2}+cu^{2}+dv^{2}\right)x^{2}}{2}}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\frac{2}{\left(1+bt^{2}+cu^{2}+dv^{2}\right)^{2}},\\ \end{align}$$

and then,

$$\begin{align} \mathcal{I}{\left(1,b,c,d\right)} &=\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}u\int_{0}^{u}\mathrm{d}v\,\frac{2}{\left(1+bt^{2}+cu^{2}+dv^{2}\right)^{2}}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{2t^{2}}{\left(1+bt^{2}+ct^{2}x^{2}+dt^{2}y^{2}\right)^{2}};~~~\small{\left[\left(u,v\right)=\left(tx,ty\right)\right]}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left(1+bt^{2}+ct^{2}x^{2}+dt^{2}y^{2}\right)^{2}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{1}\mathrm{d}t\,\frac{2t^{2}}{\left[1+\left(b+cx^{2}+dy^{2}\right)t^{2}\right]^{2}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\bigg{[}\frac{\arctan{\left(\sqrt{b+cx^{2}+dy^{2}}\right)}}{\left(b+cx^{2}+dy^{2}\right)^{3/2}}\\ &~~~~~-\frac{1}{\left(b+cx^{2}+dy^{2}\right)\left(1+b+cx^{2}+dy^{2}\right)}\bigg{]}\\ &=\int_{0}^{1}\mathrm{d}x\,\mathcal{J}{\left(b+cx^{2},d,x\right)}\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}-\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}\right].\tag{4}\\ \end{align}$$

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Define the auxiliary functions $\mathcal{F}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ and $\mathcal{G}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ by the respective integrals

$$\begin{align} \mathcal{F}{\left(b,c,d\right)} &:=\int_{0}^{1}\mathrm{d}x\,\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}\tag{5a}\\ \end{align}$$

and

$$\begin{align} \mathcal{G}{\left(b,c,d\right)} &:=\int_{0}^{1}\mathrm{d}x\,\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}.\tag{5b}\\ \end{align}$$

Suppose $\left(b,c,d\right)\in\mathbb{R}_{>0}^{3}$, and set $p:=\sqrt{\frac{b}{c}}\land q:=\sqrt{\frac{1+b}{c}}\land r:=\sqrt{\frac{d}{c}}$. Next, noting that $0<p<q$, set $s:=\frac{p}{q}\land z:=q^{-1}$. We then obtain the following expressions for $\mathcal{G}$ and $\mathcal{F}$:

$$\begin{align} \mathcal{G}{\left(b,c,d\right)} &=\int_{0}^{1}\mathrm{d}x\,\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}\\ &=\frac{1}{c\sqrt{cd}}\int_{0}^{1}\mathrm{d}x\,\frac{\arctan{\left(\frac{rx}{\sqrt{q^{2}+x^{2}}}\right)}}{\left(p^{2}+x^{2}\right)\sqrt{q^{2}+x^{2}}}\\ &=\frac{1}{c\sqrt{cd}}\int_{0}^{q^{-1}}\mathrm{d}y\,\frac{q\arctan{\left(\frac{rqy}{\sqrt{q^{2}+q^{2}y^{2}}}\right)}}{\left(p^{2}+q^{2}y^{2}\right)\sqrt{q^{2}+q^{2}y^{2}}};~~~\small{\left[x=qy\right]}\\ &=\frac{1}{c\sqrt{cd}}\int_{0}^{q^{-1}}\mathrm{d}y\,\frac{\arctan{\left(\frac{ry}{\sqrt{1+y^{2}}}\right)}}{\left(p^{2}+q^{2}y^{2}\right)\sqrt{1+y^{2}}}\\ &=\frac{1}{\left(1+b\right)\sqrt{cd}}\int_{0}^{z}\mathrm{d}y\,\frac{\arctan{\left(\frac{ry}{\sqrt{1+y^{2}}}\right)}}{\left(s^{2}+y^{2}\right)\sqrt{1+y^{2}}}\\ &=\frac{1}{\left(1+b\right)\sqrt{cd}}\int_{0}^{\frac{z}{\sqrt{1+z^{2}}}}\mathrm{d}t\,\frac{\arctan{\left(rt\right)}}{s^{2}+\left(1-s^{2}\right)t^{2}};~~~\small{\left[\frac{y}{\sqrt{1+y^{2}}}=t\right]}\\ &=\frac{1}{\left(1+b\right)\sqrt{cd}}\int_{0}^{\frac{rz}{\sqrt{1+z^{2}}}}\mathrm{d}u\,\frac{r\arctan{\left(u\right)}}{r^{2}s^{2}+\left(1-s^{2}\right)u^{2}};~~~\small{\left[rt=u\right]}\\ &=\int_{0}^{\sqrt{\frac{d}{1+b+c}}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}},\\ \end{align}$$

and

$$\begin{align} \mathcal{F}{\left(b,c,d\right)} &=\int_{0}^{1}\mathrm{d}x\,\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2x\arctan{\left(\sqrt{b+\left(c+d\right)x^{2}}\right)}}{2\left(b+cx^{2}\right)\sqrt{b+\left(c+d\right)x^{2}}}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{\arctan{\left(\sqrt{b+\left(c+d\right)y}\right)}}{2\left(b+cy\right)\sqrt{b+\left(c+d\right)y}};~~~\small{\left[x^{2}=y\right]}\\ &=\int_{b}^{b+c+d}\mathrm{d}t\,\frac{\arctan{\left(\sqrt{t}\right)}}{2\left(bd+ct\right)\sqrt{t}};~~~\small{\left[y=\frac{t-b}{c+d}\right]}\\ &=\int_{\sqrt{b}}^{\sqrt{b+c+d}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}};~~~\small{\left[\sqrt{t}=u\right]}.\\ \end{align}$$

Hence, we can express $\mathcal{I}$ as

$$\begin{align} \mathcal{I}{\left(1,b,c,d\right)} &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{x\arctan{\left(\sqrt{b+cx^{2}+dx^{2}}\right)}}{\left(b+cx^{2}\right)\sqrt{b+cx^{2}+dx^{2}}}-\frac{\arctan{\left(\frac{x\sqrt{d}}{\sqrt{1+b+cx^{2}}}\right)}}{\left(b+cx^{2}\right)\sqrt{d\left(1+b+cx^{2}\right)}}\right]\\ &=\mathcal{F}{\left(b,c,d\right)}-\mathcal{G}{\left(b,c,d\right)}\\ &=\int_{\sqrt{b}}^{\sqrt{b+c+d}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}}-\int_{0}^{\sqrt{\frac{d}{1+b+c}}}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{bd+cu^{2}}\\ &=\frac{1}{\sqrt{bcd}}\int_{\sqrt{\frac{c}{d}}}^{\sqrt{\frac{(b+c+d)c}{bd}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}\\ &~~~~~-\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{c}{b(1+b+c)}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}};~~~\small{\left[u=x\sqrt{\frac{bd}{c}}\right]}\\ &=\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{(b+c+d)c}{bd}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}-\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{c}{d}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}\\ &~~~~~-\frac{1}{\sqrt{bcd}}\int_{0}^{\sqrt{\frac{c}{b(1+b+c)}}}\mathrm{d}x\,\frac{\arctan{\left(x\sqrt{\frac{bd}{c}}\right)}}{1+x^{2}}.\tag{6}\\ \end{align}$$

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Finally, it can be shown (see Appendix) that the following integration formula holds for all $\left(p,z\right)\in\mathbb{R}_{>0}^{2}$:

$$\int_{0}^{z}\mathrm{d}x\,\frac{2\arctan{\left(px\right)}}{1+x^{2}}=\arctan^{2}{\left(z\right)}+\operatorname{Li}_{2}{\left(-\frac{1-p}{1+p}\right)}-\operatorname{Li}_{2}{\left(\frac{1-p}{1+p},\pi-2\arctan{\left(z\right)}\right)},\tag{7}$$

where the two-variable variant of the dilogarithm is defined by the integral representation

$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$

Since each of the three remaining integrals in the last line of $(6)$ above can be evaluated with formula $(7)$, this in principle completes the derivation. I don't see much point in going through the tedium of actually writing out the explicit expression though.

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Appendix.

Define the function $\mathcal{K}:\mathbb{R}_{>0}^{2}\rightarrow\mathbb{R}$ via the definite integral

$$\mathcal{K}{\left(a,z\right)}:=\int_{0}^{z}\mathrm{d}x\,\frac{\arctan{\left(ax\right)}}{1+x^{2}}.$$

In the special case $a=1$ the integral is elementary, and we have

$$\mathcal{K}{\left(1,z\right)}=\int_{0}^{z}\mathrm{d}x\,\frac{\arctan{\left(x\right)}}{1+x^{2}}=\frac12\arctan^{2}{\left(z\right)};~~~\small{z\in\mathbb{R}_{>0}}.$$

Suppose $\left(a,z\right)\in\mathbb{R}_{>0}^{2}$, and assume $a\neq1$. Then, $-1<\frac{1-a}{1+a}<1\land\frac{1-a}{1+a}\neq0$.

Set $\omega:=2\arctan{\left(z\right)}$, and note that $0<\omega<\pi\land z=\tan{\left(\frac{\omega}{2}\right)}$.

$$\begin{align} \mathcal{K}{\left(a,z\right)} &=\int_{0}^{z}\mathrm{d}x\,\frac{\arctan{\left(ax\right)}}{1+x^{2}}\\ &=\int_{0}^{z}\mathrm{d}x\,\frac{1}{1+x^{2}}\int_{0}^{a}\mathrm{d}t\,\frac{x}{1+x^{2}t^{2}}\\ &=\int_{0}^{z}\mathrm{d}x\int_{0}^{a}\mathrm{d}t\,\frac{x}{\left(1+x^{2}\right)\left(1+t^{2}x^{2}\right)}\\ &=\int_{0}^{a}\mathrm{d}t\int_{0}^{z}\mathrm{d}x\,\frac{x}{\left(1+x^{2}\right)\left(1+t^{2}x^{2}\right)}\\ &=\frac12\int_{0}^{a}\mathrm{d}t\int_{0}^{z^{2}}\mathrm{d}y\,\frac{1}{\left(1+y\right)\left(1+t^{2}y\right)};~~~\small{\left[x^{2}=y\right]}\\ &=\frac12\int_{0}^{a}\mathrm{d}t\int_{0}^{z^{2}}\mathrm{d}y\,\frac{d}{dy}\left[\frac{\ln{\left(1+y\right)}-\ln{\left(1+t^{2}y\right)}}{1-t^{2}}\right]\\ &=\frac12\int_{0}^{a}\mathrm{d}t\,\frac{\ln{\left(1+z^{2}\right)}-\ln{\left(1+z^{2}t^{2}\right)}}{1-t^{2}}\\ &=-\frac12\int_{0}^{a}\mathrm{d}t\,\frac{\ln{\left(\frac{1+z^{2}t^{2}}{1+z^{2}}\right)}}{1-t^{2}}\\ &=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\ln{\left(\frac{1+z^{2}\left(\frac{1-x}{1+x}\right)^{2}}{1+z^{2}}\right)};~~~\small{\left[t=\frac{1-x}{1+x}\right]}\\ &=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\ln{\left(\frac{\left(1+x\right)^{2}+z^{2}\left(1-x\right)^{2}}{\left(1+z^{2}\right)\left(1+x\right)^{2}}\right)}\\ &=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\ln{\left(\frac{1+2\left(\frac{1-z^{2}}{1+z^{2}}\right)x+x^{2}}{\left(1+x\right)^{2}}\right)}\\ &=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{1}{2x}\left[\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}-2\ln{\left(1+x\right)}\right]\\ &=-\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}+\frac12\int_{\frac{1-a}{1+a}}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{x}\\ &=-\frac12\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}+\frac12\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}\\ &~~~~~-\frac12\operatorname{Li}_{2}{\left(-1\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}\\ &=+\frac12\operatorname{Li}_{2}{\left(-1\right)}+\frac18\omega^{2}+\frac12\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}x\,\frac{\ln{\left(1+2x\cos{\left(\omega\right)}+x^{2}\right)}}{2x}\\ &~~~~~-\frac12\operatorname{Li}_{2}{\left(-1\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}\\ &=\frac12\arctan^{2}{\left(z\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}\\ &~~~~~-\frac12\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}x\,\frac{(-1)\ln{\left(1-2x\cos{\left(\pi-\omega\right)}+x^{2}\right)}}{2x}\\ &=\frac12\arctan^{2}{\left(z\right)}+\frac12\operatorname{Li}_{2}{\left(-\frac{1-a}{1+a}\right)}-\frac12\operatorname{Li}_{2}{\left(\frac{1-a}{1+a},\pi-\omega\right)}.\blacksquare\\ \end{align}$$

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