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$$\int_{-\infty}^\infty {{\cos(mx)}\over(x^2+a^2)(x^2+b^2)}dx={\pi(ae^{-mb}-be^{-ma})\over ab(a^2-b^2)}$$

where $a,b,m>0$ and $a$ is not equal to $b$.

I already know that $\int_{-\infty}^\infty {{1} \over(x^2+a^2)(x^2+b^2)}dx={\pi\over ab(a+b)}$.

I'm wondering if I can use this fact at all.

I'm not sure how to deal with the $\cos(mx)$ term.

Any solutions or help is greatly appreciated.

Olivier Oloa
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  • That known formula is a special case with $m=0$. As you wand to generalize rather than specialize, it won't be so helpful. –  Apr 22 '16 at 14:58
  • 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. – heropup Apr 22 '16 at 17:44

1 Answers1

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Hint. One may write, for $a^2\neq b^2$, $$ \int_{-\infty}^{+\infty}\frac{\cos(mx)\:dx}{\left(x^2+a^2\right)\left(x^2+b^2\right)}=\frac{-1}{\left(a^2-b^2\right)}\left(\int_{-\infty}^{+\infty}\frac{\cos(mx)}{\left(x^2+a^2\right)}\:dx-\int_{-\infty}^{+\infty}\frac{\cos(mx)}{\left(x^2+b^2\right)}\:dx\right) $$ then use the classic evaluation

$$ \int_{-\infty}^{+\infty}\frac{\cos(mx)}{x^2+a^2}dx=\frac{\pi}{a}\:e^{-a m},\quad a>0,\,m>0, $$

which is proved for example in the accepted answer here.

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Olivier Oloa
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