Integral $\int_{0}^{\infty}\frac{dy}{1+y^{2}}\log\left(\sqrt{1+y^{2}}+\sqrt{x+y^{2}}\right)$

Question

I need help finding an analytical expression for the integral $$I(x)=\int_{0}^{\infty}\frac{dy}{1+y^{2}}\log\left(\sqrt{1+y^{2}}+\sqrt{x+y^{2}}\right), $$ where $0<x<1$. The expression can be written using polylog, elliptic function or any other known integral functions...

Answer 1

OK, I'm not super happy with what I have, but I'm posting in case it helps others find a better formula.

$$\boxed{I(x)=\pi\log 2 -\frac 1 2 \int_x^1 \frac{K(\sqrt{1-t})-\frac \pi 2}{1-t}dt=\pi\log 2-\frac {\pi} 4 \sum_{n\geq 1}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2\frac{(1-x)^{n+1}}{n+1}}$$ where we identify the complete elliptic integral of the first kind $$K(k)=\int_0^{\frac \pi 2} \frac{d\theta}{\sqrt{1 -k^2\sin^2\theta}}$$ Thre might be a way to get another form using hypergeometric functions, but I couldn't. Anyway, here's what I did:

$$\begin{split} I^\prime(x)&= \int_0^{+\infty} \frac{dy}{2(y^2+1)\sqrt{x+y^2}\left(\sqrt{x+y^2}+\sqrt{y^2+1}\right)}\\ &= \int_0^{+\infty} \frac{\sqrt{x+y^2}-\sqrt{y^2+1}}{2(y^2+1)\sqrt{x+y^2}(x-1)}dy\\ &=\frac 1 {2(x-1)}\left(\int_0^{+\infty} \frac{1}{y^2+1}dy - \int_0^{+\infty} \frac{dy}{\sqrt{(y^2+1)(x+y^2)}} \right)\\ &= \frac 1 {2(x-1)}\left(\frac \pi 2 - \int_0^{\frac \pi 2} \frac{d\theta}{\sqrt{\cos^2\theta + x\sin^2\theta}} \right) \text { with }\theta=\arctan \frac y {\sqrt x}\\ &= \frac 1 {2(x-1)}\left(\frac \pi 2 - \int_0^{\frac \pi 2} \frac{d\theta}{\sqrt{1 - (1-x)\sin^2\theta}} \right)\\ &= \frac 1 {2(x-1)}\left(\frac \pi 2 - K(\sqrt{1-x})\right) \end{split}$$ Using the power series for $K$, $$K(\sqrt{1-x})=\frac \pi 2 \sum_{n\geq 0}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2(1-x)^n$$ we obtain $$\begin{split} I(x)-I(1)&=\frac {\pi} 4 \sum_{n\geq 1}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2\int_1^x (1-t)^ndt\\ &=-\frac {\pi} 4 \sum_{n\geq 1}\left( \frac{(2n)!}{2^{2n}(n!)^2}\right)^2\frac{(1-x)^{n+1}}{n+1} \end{split}$$ Finally $$\begin{split} I(1)&=\int_0^{+\infty}\frac{\log(2\sqrt{1+y^2})}{1+y^2}dy\\ &=\frac \pi 2 \log 2+\int_0^{+\infty}\frac{\log(\sqrt{1+y^2})}{1+y^2}dy\\ &=\frac \pi 2 \log 2-\int_0^{\frac \pi 2}\log(\cos \theta)d\theta \\ &= \pi \log 2 \end{split}$$ where we have used this result.