Determine whether $\int_{0}^{\infty}\frac{\log(1+\left(ax^{2}\right))}{1+x^{2}}dx$ converges

Question

I know it's a silly question but I cannot find any map to define whether is convergent or not this improper integral $\int_{0}^{\infty}\frac{\log(1+\left(ax^{2}\right))}{1+x^{2}}dx$. Any ideas?

Answer 1

You could make some assumptions to show whether or not this is convergent, for example: $$\int_0^\infty\frac{\log(1+ax^2)}{1+x^2}dx\approx\int_0^\infty\frac{\log(ax^2)}{1+x^2}dx=\int_0^\infty\frac{\log(a)+2\log(x)}{1+x^2}dx$$ and as $x\to\infty$ this is of the form $\frac{\log(x)}{1+x^2}$. The strength of the convergence of this determines the convergence of the integral

Answer 2

Feynman's trick (fifferentiation under the integral sign) would be very useful for your problem.

$$I=\int_{0}^{\infty}\frac{\log(1+ax^{2})}{1+x^{2}}\,dx\implies \frac {dI}{da}=\int_{0}^{\infty}\frac{x^2}{\left(1+x^2\right) \left(1+a x^2\right)}\,dx$$ Using partial fraction decomposition $$\frac{x^2}{\left(x^2+1\right) \left(a x^2+1\right)}=\frac{1}{(a-1) \left(1+x^2\right)}-\frac{1}{(a-1) \left(1+a x^2\right)}$$ $$\int\frac{x^2}{\left(x^2+1\right) \left(a x^2+1\right)}\,dx=\frac{1}{a-1}\left(\tan ^{-1}(x)-\frac{\tan ^{-1}\left(x\sqrt{a} \right)}{\sqrt{a}} \right)$$ Assuming $a>0$, then $$\frac {dI}{da}=\frac{\pi }{2 \left(a+\sqrt{a}\right)}\implies I=\pi \log \left(1+\sqrt{a}\right)$$