Evaluate $\int^\infty_0\frac{x^{s-1}\arctan{x}}{\sqrt{1+x^2}}\ dx$

Question

I was playing around with this integral and thought it may be of interest to some: $$\int^\infty_0\frac{x^{s-1}\arctan{x}}{\sqrt{1+x^2}}\ dx=\frac{1}{2\cos\left(\frac{\pi s}{2}\right)}\int^1_0\frac{x^{-s}\ln\left(\frac{1+x}{1-x}\right)}{\sqrt{1-x^2}}\ dx=\frac{\sqrt\pi}{2\cos\left(\frac{\pi s}{2}\right)}\frac{\Gamma\left(1-\tfrac s2\right)}{\Gamma\left(\tfrac{3-s}2\right)}{}_3F_2\left(\tfrac12, 1-\tfrac s2,1;\tfrac 32,\tfrac{3-s}2;1\right)$$ Edit: After a series of further manipulations I managed to obtain the identity $$\int^\infty_0\frac{J_0(x)\left[\left(C\left(\sqrt{\frac{2x}\pi}\right)-\frac12\right)^2+\left(S\left(\sqrt{\frac{2x}\pi}\right)-\frac12\right)^2\right]}{\sqrt{x}}\ dx=\frac{\Gamma^2\left(\frac{1}{4}\right)}{8\pi}$$ where $J_0(x)$ is the zeroth-order Bessel function of the first kind and $C$ and $S$ are Fresnel integrals.

Answer 1

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Define the function $\mathcal{I}:(-1,1)\rightarrow\mathbb{R}$ via the improper integral

$$\mathcal{I}{\left(s\right)}:=\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{s-1}\arctan{\left(x\right)}}{\sqrt{1+x^{2}}}.$$

Recall that we can represent the real arctangent function by either of the integrals

$$\arctan{\left(x\right)}=\int_{0}^{x}\mathrm{d}t\,\frac{1}{1+t^{2}}=\int_{0}^{1}\mathrm{d}y\,\frac{x}{1+x^{2}y^{2}}.$$

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Suppose $s\in(-1,1)$.

$$\begin{align} \mathcal{I}{\left(s\right)} &=\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{s-1}\arctan{\left(x\right)}}{\sqrt{1+x^{2}}}\\ &=\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{s-1}}{\sqrt{1+x^{2}}}\int_{0}^{1}\mathrm{d}y\,\frac{x}{1+x^{2}y^{2}}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{x^{s-1}}{\sqrt{1+x^{2}}}\cdot\frac{x}{1+x^{2}y^{2}}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{s}}{\left(1+x^{2}y^{2}\right)\sqrt{1+x^{2}}}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}w\,\frac{w^{1-s}}{\left(w^{2}+y^{2}\right)\sqrt{w^{2}+1}};~~~\small{\left[x=w^{-1}\right]}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}v\,\frac{v^{-s/2}}{2\left(v+y^{2}\right)\sqrt{v+1}};~~~\small{\left[w=\sqrt{v}\right]}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}v\,\frac{\left(\frac{v}{1+v}\right)^{-s/2}\left(\frac{1}{1+v}\right)^{s/2+1/2}}{2\left(v+y^{2}\right)}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}u\,\frac{u^{-s/2}\left(1-u\right)^{s/2-3/2}}{2\left(\frac{u}{1-u}+y^{2}\right)};~~~\small{\left[v=\frac{u}{1-u}\right]}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}u\,\frac{u^{-s/2}\left(1-u\right)^{s/2-1/2}}{2\left[u+y^{2}\left(1-u\right)\right]}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}u\,\frac{u^{-s/2}\left(1-u\right)^{s/2-1/2}}{2\left[y^{2}+\left(1-y^{2}\right)u\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{1}{2y^{2}}\int_{0}^{1}\mathrm{d}u\,\frac{u^{-s/2}\left(1-u\right)^{s/2-1/2}}{\left[1-\left(1-y^{-2}\right)u\right]}.\\ \end{align}$$

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Recall Euler's integral formula for the Gauss hypergeometric function, which states that for real parameters and variables we have

$$\int_{0}^{1}\mathrm{d}t\,\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-zt\right)^{a}}=\operatorname{B}{\left(b,c-b\right)}\,{_2F_1}{\left(a,b;c;z\right)};~~~\small{0<b<c\land z<1}.$$

There are similar integration formulas for higher order hypergeometric functions. For $z\le1\land0<p<q\land0<c+q-a-b-p$, we have

$$\int_{0}^{1}\mathrm{d}t\,t^{p-1}\left(1-t\right)^{q-p-1}\,{_2F_1}{\left(a,b;c;zt\right)}=\operatorname{B}{\left(p,q-p\right)}\,{_3F_2}{\left(a,b,p;c,q;z\right)}.$$

It follows that for $-1<s<2\land0<y$ we have

$$\int_{0}^{1}\mathrm{d}u\,\frac{u^{-s/2}\left(1-u\right)^{s/2-1/2}}{\left[1-\left(1-y^{-2}\right)u\right]}=\operatorname{B}{\left(1-\frac{s}{2},\frac{s+1}{2}\right)}\,{_2F_1}{\left(1,1-\frac{s}{2};\frac32;1-y^{-2}\right)}.$$

It can be shown that the Gauss hypergeometric function obeys the following functional relation, known as Pfaff's transformation:

$${_2F_1}{\left(a,b;c;z\right)}=\left(1-z\right)^{-b}\,{_2F_1}{\left(c-a,b;c;\frac{z}{z-1}\right)};~~~\small{0<b<c\land z<1}.$$

Then,

$${_2F_1}{\left(1,1-\frac{s}{2};\frac32;1-y^{-2}\right)}=\left(y^{2}\right)^{1-\frac{s}{2}}\,{_2F_1}{\left(\frac12,1-\frac{s}{2};\frac32;1-y^{2}\right)};~~~\small{-1<s<2\land0<y}.$$

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Continuing with our main calculation,

$$\begin{align} \mathcal{I}{\left(s\right)} &=\int_{0}^{1}\mathrm{d}y\,\frac{1}{2y^{2}}\int_{0}^{1}\mathrm{d}u\,\frac{u^{-s/2}\left(1-u\right)^{s/2-1/2}}{\left[1-\left(1-y^{-2}\right)u\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{1}{2y^{2}}\operatorname{B}{\left(1-\frac{s}{2},\frac{s+1}{2}\right)}\,{_2F_1}{\left(1,1-\frac{s}{2};\frac32;1-y^{-2}\right)}\\ &=\frac12\operatorname{B}{\left(1-\frac{s}{2},\frac{s+1}{2}\right)}\int_{0}^{1}\mathrm{d}y\,\left(y^{2}\right)^{-s/2}\,{_2F_1}{\left(\frac12,1-\frac{s}{2};\frac32;1-y^{2}\right)}\\ &=\frac12\operatorname{B}{\left(1-\frac{s}{2},\frac{s+1}{2}\right)}\int_{0}^{1}\mathrm{d}u\,\frac{u^{-s/2}}{2\sqrt{u}}\,{_2F_1}{\left(\frac12,1-\frac{s}{2};\frac32;1-u\right)};~~~\small{\left[y=\sqrt{u}\right]}\\ &=\frac14\operatorname{B}{\left(1-\frac{s}{2},\frac{s+1}{2}\right)}\int_{0}^{1}\mathrm{d}u\,u^{-s/2-1/2}\,{_2F_1}{\left(\frac12,1-\frac{s}{2};\frac32;1-u\right)}\\ &=\frac14\operatorname{B}{\left(1-\frac{s}{2},\frac{s+1}{2}\right)}\int_{0}^{1}\mathrm{d}t\,\left(1-t\right)^{-s/2-1/2}\,{_2F_1}{\left(\frac12,1-\frac{s}{2};\frac32;t\right)};~~~\small{\left[u=1-t\right]}\\ &=\frac14\operatorname{B}{\left(1-\frac{s}{2},\frac{s+1}{2}\right)}\operatorname{B}{\left(1,\frac{1-s}{2}\right)}\,{_3F_2}{\left(\frac12,1-\frac{s}{2},1;\frac32,\frac{3-s}{2};1\right)}\\ &=\frac14\cdot\frac{\Gamma{\left(1-\frac{s}{2}\right)}\,\Gamma{\left(\frac{s+1}{2}\right)}}{\Gamma{\left(\frac32\right)}}\cdot\frac{\Gamma{\left(1\right)}\,\Gamma{\left(\frac{1-s}{2}\right)}}{\Gamma{\left(\frac{3-s}{2}\right)}}\,{_3F_2}{\left(\frac12,1-\frac{s}{2},1;\frac32,\frac{3-s}{2};1\right)}\\ &=\frac{\Gamma{\left(\frac{s+1}{2}\right)}\,\Gamma{\left(\frac{1-s}{2}\right)}}{2\,\Gamma{\left(\frac12\right)}}\cdot\frac{\Gamma{\left(1-\frac{s}{2}\right)}}{\Gamma{\left(\frac{3-s}{2}\right)}}\,{_3F_2}{\left(\frac12,1-\frac{s}{2},1;\frac32,\frac{3-s}{2};1\right)}\\ &=\frac{\pi\csc{\left(\pi\left(\frac{s+1}{2}\right)\right)}}{2\,\Gamma{\left(\frac12\right)}}\cdot\frac{\Gamma{\left(1-\frac{s}{2}\right)}}{\Gamma{\left(\frac{3-s}{2}\right)}}\,{_3F_2}{\left(\frac12,1-\frac{s}{2},1;\frac32,\frac{3-s}{2};1\right)}\\ &=\frac{\sqrt{\pi}}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\cdot\frac{\Gamma{\left(1-\frac{s}{2}\right)}}{\Gamma{\left(\frac{3-s}{2}\right)}}\,{_3F_2}{\left(\frac12,1-\frac{s}{2},1;\frac32,\frac{3-s}{2};1\right)}\\ &=\frac{1}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\,\operatorname{B}{\left(1-\frac{s}{2},\frac12\right)}\,{_3F_2}{\left(\frac12,1-\frac{s}{2},1;\frac32,\frac{3-s}{2};1\right)},\\ \end{align}$$

which is the proposed hypergeometric form. This is interesting (and what prompted me to want to work on this problem!) because a more straightforward route of obtaining a hypergeometric form yields one that isn't obviously equal to the form above: for $s\in(-1,1)$ we have

$$\begin{align} \mathcal{I}{\left(s\right)} &=\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{s-1}\arctan{\left(x\right)}}{\sqrt{1+x^{2}}}\\ &=\int_{0}^{\infty}\mathrm{d}y\,\frac{y^{(s-1)/2}\arctan{\left(\sqrt{y}\right)}}{2\sqrt{y}\sqrt{1+y}};~~~\small{\left[x=\sqrt{y}\right]}\\ &=\frac12\int_{0}^{\infty}\mathrm{d}y\,y^{s/2-1}\left(1+y\right)^{-1/2}\arctan{\left(\sqrt{y}\right)}\\ &=\frac12\int_{0}^{\infty}\mathrm{d}y\,\left(\frac{y}{1+y}\right)^{s/2-1}\left(1-\frac{y}{1+y}\right)^{(3-s)/2}\arctan{\left(\sqrt{y}\right)}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,t^{s/2-1}\left(1-t\right)^{-s/2-1/2}\arctan{\left(\sqrt{\frac{t}{1-t}}\right)};~~~\small{\left[y=\frac{t}{1-t}\right]}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,t^{s/2-1}\left(1-t\right)^{-s/2-1/2}\arcsin{\left(\sqrt{t}\right)}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,t^{s/2-1/2}\left(1-t\right)^{-s/2-1/2}\,\frac{\arcsin{\left(\sqrt{t}\right)}}{\sqrt{t}}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,t^{s/2-1/2}\left(1-t\right)^{-s/2-1/2}\,{_2F_1}{\left(\frac12,\frac12;\frac32;t\right)}\\ &=\frac12\operatorname{B}{\left(\frac{s+1}{2},\frac{1-s}{2}\right)}\,{_3F_2}{\left(\frac12,\frac12,\frac{s+1}{2};\frac32,1;1\right)}\\ &=\frac{\pi}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\,{_3F_2}{\left(\frac12,\frac12,\frac{s+1}{2};\frac32,1;1\right)}.\\ \end{align}$$

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Finally, we can verify the conjectured integral equality from the integral representation of the hypergeometric form we found. For $s\in(-1,1)$,

$$\begin{align} \mathcal{I}{\left(s\right)} &=\frac{1}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\,\operatorname{B}{\left(1-\frac{s}{2},\frac12\right)}\,{_3F_2}{\left(\frac12,1-\frac{s}{2},1;\frac32,\frac{3-s}{2};1\right)}\\ &=\frac{1}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\,\operatorname{B}{\left(1-\frac{s}{2},\frac12\right)}\,{_3F_2}{\left(\frac12,1,1-\frac{s}{2};\frac32,\frac{3-s}{2};1\right)}\\ &=\frac{1}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\int_{0}^{1}\mathrm{d}t\,t^{-s/2}\left(1-t\right)^{-1/2}\,{_2F_1}{\left(\frac12,1;\frac32;t\right)}\\ &=\frac{1}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\int_{0}^{1}\mathrm{d}t\,t^{-s/2}\left(1-t\right)^{-1/2}\,{_2F_1}{\left(1,\frac12;\frac32;t\right)}\\ &=\frac{1}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\int_{0}^{1}\mathrm{d}t\,t^{-s/2}\left(1-t\right)^{-1/2}\,\frac{1}{\operatorname{B}{\left(\frac12,1\right)}}\int_{0}^{1}\mathrm{d}u\,\frac{u^{-1/2}}{\left(1-tu\right)}\\ &=\frac{1}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\int_{0}^{1}\mathrm{d}t\,\frac{t^{-s/2}}{2\sqrt{1-t}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\left(1-tu\right)\sqrt{u}}\\ &=\frac{1}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\int_{0}^{1}\mathrm{d}x\,\frac{x^{1-s}}{\sqrt{1-x^{2}}}\int_{0}^{1}\mathrm{d}u\,\frac{1}{\left(1-x^{2}u\right)\sqrt{u}};~~~\small{\left[t=x^{2}\right]}\\ &=\frac{1}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\int_{0}^{1}\mathrm{d}x\,\frac{x^{-s}}{\sqrt{1-x^{2}}}\int_{0}^{1}\mathrm{d}y\,\frac{2x}{1-x^{2}y^{2}};~~~\small{\left[u=y^{2}\right]}\\ &=\frac{1}{2\cos{\left(\frac{\pi\,s}{2}\right)}}\int_{0}^{1}\mathrm{d}x\,\frac{x^{-s}}{\sqrt{1-x^{2}}}\ln{\left(\frac{1+x}{1-x}\right)}.\blacksquare\\ \end{align}$$

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