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: $$ \frac12\int_0^\infty \frac1{(s+1)^2}\mathrm ds=\left[-\frac1{2(s+1)}\right]^\infty_0=\frac12. $$

  • $\begingroup$ How did I miss that ??? Thank you very much ! $\endgroup$ – dzhqjk Feb 27 '19 at 21:49
  • $\begingroup$ I hope it was helpful. $\endgroup$ – Song Feb 27 '19 at 21:51
  • $\begingroup$ Yes thank you very much. Actually the integral is not defined for $x = 1$. I should say $x > 0$ instead. $\endgroup$ – dzhqjk Feb 27 '19 at 22:03

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