Difficult logarithmic integral: $\int_0^\Lambda r^{d - 1}\log(1 + a\sqrt{r^2 + m_1^2} + b\sqrt{r^2 + m_2^2}) dr$

Question

I'm working on QFT's, and encountered the following integral \begin{equation} \begin{aligned} I = \int_0^\Lambda r^{d - 1}\log(1 + a\sqrt{r^2 + m_1^2} + b\sqrt{r^2 + m_2^2}) dr \ , \end{aligned} \end{equation} where $a, b, m_1, m_2$ are real, non-zero parameters all different from each other, and $\Lambda > 1$. (I expect the integral to be divergent in the $\Lambda\rightarrow\infty$ limit.) I'm wondering if anyone here can help me perform this integration. I believe it might be useful to differentiate w.r.t. $a$ \begin{equation} \begin{aligned} \frac{\partial I}{\partial a} = \int_0^\Lambda \frac{r^{d - 1}\sqrt{r^2 + m_1^2}}{1 + a\sqrt{r^2 + m_1^2} + b\sqrt{r^2 + m_2^2}} dr \ . \end{aligned} \end{equation} I tried to do these integrations in Mathematica, but it wasn't able to do it. That said, if you know a program that can do these integrals that's good enough. Moreover, finding a primitive function is probably also good enough.

I'm grateful for any help!

Answer 1

Of course, with $\Lambda$ there, you are just asking for the indefinite integral.

Why do you want it in closed form? Don't you think leaving it in integral form is more useful?

With $d=a=b=1$, Maple finds an answer. It is long. It involves arctan, atanh, elliptic integrals $E$, $\Pi$, and $F$. Here it is an Maple notation:

1/2/(-m[1]^2+(2*m[2]+2)*m[1]-(m[2]-1)^2)^(1/2)*(1/4*I*(m[2]^(1/2)*m[1]^2-2*m[2] ^(3/2)*m[1]-m[2]^(1/2)+m[2]^(5/2))*(-m[1]^2+(2*m[2]+2)*m[1]-(m[2]-1)^2)^(1/2)* EllipticPi(I*r/m[1]^(1/2),-4*m[1]/(m[1]^2+(-2*m[2]-2)*m[1]+(m[2]-1)^2),1/m[2]^( 1/2)*m[1]^(1/2))+(-1/4*m[2]*(arctanh((r*(m[1]^2+(-2*m[2]-2)*m[1]+(m[2]-1)^2)^(1 /2)-2*m[1])/(r^2+m[1])^(1/2)/(m[1]-m[2]+1))+arctanh((r*(m[1]^2+(-2*m[2]-2)*m[1] +(m[2]-1)^2)^(1/2)+2*m[1])/(r^2+m[1])^(1/2)/(m[1]-m[2]+1))-arctanh((r*(m[1]^2+( -2*m[2]-2)*m[1]+(m[2]-1)^2)^(1/2)-2*m[2])/(r^2+m[2])^(1/2)/(m[1]-m[2]-1))- arctanh((r*(m[1]^2+(-2*m[2]-2)*m[1]+(m[2]-1)^2)^(1/2)+2*m[2])/(r^2+m[2])^(1/2)/ (m[1]-m[2]-1)))*(m[1]^2+(-2*m[2]-2)*m[1]+(m[2]-1)^2)^(1/2)+2*ln(1+(r^2+m[1])^(1 /2)+(r^2+m[2])^(1/2))*r*m[2]+((-I+1/2*I*m[1])*m[2]^(3/2)-1/4*I*m[2]^(5/2)+(1/4* I-1/4*I*m[1]^2)*m[2]^(1/2))*EllipticF(I*r/m[1]^(1/2),1/m[2]^(1/2)*m[1]^(1/2))+I *EllipticE(I*r/m[1]^(1/2),1/m[2]^(1/2)*m[1]^(1/2))*m[2]^(3/2)-m[2]*((-1/2*m[1]+ 1/2*m[2]-1/2)*ln(r+(r^2+m[1])^(1/2))+(1/2*m[1]-1/2*m[2]-1/2)*ln(r+(r^2+m[2])^(1 /2))+r))*(-m[1]^2+(2*m[2]+2)*m[1]-(m[2]-1)^2)^(1/2)-1/2*(m[2]^2+(-2*m[1]-2)*m[2 ]+(m[1]-1)^2)*m[2]*arctan(2*r/(-m[1]^2+(2*m[2]+2)*m[1]-(m[2]-1)^2)^(1/2)))/m[2]

Note m[1] and m[2] are $m_1, m_2$. Assumptions were: $m_1 > m_2+1, m_2 > 0$.

If I try $d=2$, Maple does not solve it.
If I try $d=1$ but $a,b$ merely positive, Maple solves it. But even in Maple notation its length exceeds $10^6$.