The integral of $(1+x^2)^{-a} \ln (1+x^2)$

Question

Consider the integral $$ \int_0^{\infty} (1+x^2)^{-a} \ln (1+x^2) dx$$ where $a > 0$ is big enough. Is it expressible in terms of some known special functions (gamma function, hypergeometric function, etc.)? Thanks!

Answer 1

Use this identity:

$$B(x,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}dt$$ In your case use in addition the change of variables $x^2=t$ to get $$\int_0^\infty\frac{1}{(1+x^2)^{a}}dx=\frac{1}{2}\int_0^\infty\frac{t^{1/2-1}}{(1+t)^{a-1/2+1/2}}dt=\frac{1}{2}B\left(\frac{1}{2}, a-\frac{1}{2}\right)$$ Now derive with respect to $a$ to get your integral. After all you should get

$$\int_0^\infty\frac{\ln(1+x^2)}{(1+x^2)^a}dx=\frac{B\left(\frac{1}{2}, a-\frac{1}{2}\right)}{2}\left(\psi(a)-\psi\left(a-\frac{1}{2}\right)\right)$$

Update: $B(x,y)$ is the beta function and $\psi(x)$ is the digamma function

https://en.wikipedia.org/wiki/Beta_function

https://en.wikipedia.org/wiki/Digamma_function