Show that $$\int_0^\infty {dx\over (x^2+a^2)(x^2+b^2)}={\pi\over2ab(a+b)}$$ where $a,b>0$. I'm not sure how to simplify this. Any solutions or hints are greatly appreciated.


closed as off-topic by heropup, user296602, C. Dubussy, JKnecht, Martin R Apr 21 '16 at 21:24

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    $\begingroup$ Since January 20 all but one of the questions you have posted have lacked context. Several of your posts have been closed for this reason, yet you continue to ignore all warnings that the posting of questions without any explanation of your own thoughts or effort to solve them is not allowed on this site. $\endgroup$ – heropup Apr 21 '16 at 20:20
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    $\begingroup$ $$\frac{1}{(x^2+a^2)(x^2+b^2)}=\frac{1}{b^2-a^2}\left(\frac{1}{x^2+a^2} - \frac{1}{b^2+x^2}\right)$$ $\endgroup$ – Jack D'Aurizio Apr 21 '16 at 20:39

Fill in details: take the contour

$$C_R:=[-R,R]\cup\gamma_R\;,\;\;R>>0\;,\;\;\gamma_R:=\{z=Re^{it}\in\Bbb C\;:\;0\le t\le \pi\}\;$$

For $\;a\neq b\;$ :

and $\;R\;$ big enough as to be $\;a,b<R\;$ . Observe there are two simple poles of $\;f(z)=\cfrac1{(z^2+a^2)(z^2+b^2)}\;$ within the domain enclosed by the above contour:

$$\begin{align}&\text{Res}_{z=ai}(f)=\lim_{z\to ai}(z-ai)f(z)=\frac1{2ai(b^2-a^2)}\\{}\\ &\text{Res}_{z=bi}(f)=\lim_{z\to bi}(z-bi)f(z)=\frac1{2bi(a^2-b^2)}\end{align}$$

and from the Residue Theorem:


Now show (for example, Jordan's Lemma) that


and use the fact that the real function $\;f(x)\;$ is even to obtain the result.

For $\;a=b\;$ : we now have the function $\;f(z)=\cfrac1{(z^2+a^2)^2}=\cfrac1{(z+ai)^2(z-ai)^2}\;$ . Thus, using the same contour as above and taking a little circle $\;|z-ai|=r\;,\;\;0<r<<R\;$ , we get from Cauchy Integral Formulae

$$\oint_{C_R}f(z)dz=\oint_{|z|=r}\frac{\frac1{(z+ai)^2}}{(z-ai)^2}dz=2\pi i\left(\frac1{(z+ai)^2}\right)'_{z=ai}=$$

$$=-2\pi i\frac{2}{(2ai)^3}=\frac\pi{a^3}$$

and continue asin the first part.

  • $\begingroup$ Somebody apparently didn't like this hinted answer. I wonder why? $\endgroup$ – DonAntonio Apr 21 '16 at 20:21
  • $\begingroup$ Maybe because you must pay attention that $a$ can be equal to $b$. $\endgroup$ – C. Dubussy Apr 21 '16 at 20:28
  • $\begingroup$ @C.Dubussy Thank you. Yes, perhaps because of that, though if $\;a=b\;$ then it is even a little simpler. Yet, without saying, is hard to guess that is the reason, if indeed that is. I shall try to edit that in short time. $\endgroup$ – DonAntonio Apr 21 '16 at 20:31
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    $\begingroup$ Solid answer. +1 $\endgroup$ – Mark Viola Apr 21 '16 at 22:57

You should take the partial fraction of $\dfrac{1}{(x^2+a^2)(x^2+b^2)}$.

This way, it is easier for you to integrate the expression and you should expect the $\arctan$ function in the numerator.

Recall: $\dfrac{1}{(x^2+a^2)(x^2+b^2)}=\dfrac{Ax+B}{x^2+a^2}+\dfrac{Cx+D}{x^2+b^2}$.

  • $\begingroup$ What's wrong with this answer? It looks perfectly fine to me. (+1) back. $\endgroup$ – Jack D'Aurizio Apr 21 '16 at 21:01
  • $\begingroup$ Solid answer. +1 $\endgroup$ – Mark Viola Apr 21 '16 at 22:57
  • $\begingroup$ Two persons downvoted both answers apparently just because. If somebody tells me what is wrong, what wasn't taken into account it's fine, but just like that? Odd. $\endgroup$ – DonAntonio Apr 22 '16 at 6:52

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