# Evaluate $\int_0^{\infty} \frac{t}{(t^2+1)(t^2+x^2)} \mathrm{d}t$

Let $$x \geq0$$ : $$\int_0^{\infty} \frac{t}{(t^2+1)(t^2+x^2)} \mathrm{d}t = ?$$

A friend of mine told me that the answer can be expressed using $$\ln$$ and an inverse of a polynomial. I must say that I don't know how to proceed. I am completely stuck. There is no partial fraction decomposition here (in the real).

The denominator reminds me the derivative of $$\tan^{-1}$$ so I don't really see how $$\ln$$ is involved here. I also tried substitution but the squares makes it hard. For example when there are square we like to substitute trigonometric functions yet here it's impossible since we are integrating on $$]0,+\infty)$$ thus these subsitutions are impossible.

Thank you !

Making change of variable $$s=t^2$$, we have for $$x\ne 1$$, \begin{align*} I&=\int_0^{\infty} \frac{t}{(t^2+1)(t^2+x^2)} \mathrm{d}t\\&=\frac12\int_0^\infty \frac1{(s+1)(s+x^2)}\mathrm ds\\&=\frac1{2(x^2-1)}\int_0^\infty\left( \frac1{s+1}-\frac1{s+x^2}\right)\mathrm ds\\&=\frac1{2(x^2-1)}\left[\ln\left(\frac{s+1}{s+x^2}\right)\right]^\infty_0\\&=\frac{\ln x}{x^2-1}. \end{align*} For $$x=1$$, we have $$I =\lim_{x\to 1}\frac{\ln x}{x^2-1}=\frac12$$. We can also do it explicitly: $$\frac12\int_0^\infty \frac1{(s+1)^2}\mathrm ds=\left[-\frac1{2(s+1)}\right]^\infty_0=\frac12.$$
• Yes thank you very much. Actually the integral is not defined for $x = 1$. I should say $x > 0$ instead. – dzhqjk Feb 27 '19 at 22:03