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Does this integral seem to have a nice closed form (at least for a subset of values of $a > 0$)

$$ \int_{\sqrt{a}}^\infty \frac{y^2}{(y^2-a+1)^2} dy $$

Using a symbolic math software and for $a$ being some small integers, I end up getting different answers in terms of $\sinh^{-1}$, $\log$, $\tanh^{-1}$ and $\coth^{-1}$ depending on $a$.

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I guess one might use a partial fraction expansion, ... but is there any other way. – passerby51 Oct 16 '12 at 18:48
let $b^2 = a-1$, then the denominator $(y^2-b^2)^2$ suggests a trigonometric substitution. Different values of $a$ will depend on the sign of $b$: $b=0$ is arithmetic, and you may use $\tan$ for $b<0$ and $\sin$ for $b > 0$... – gt6989b Oct 16 '12 at 19:14
up vote 2 down vote accepted

Partial fractions does it. The partial fraction expansion of the integrand is

$$ \frac{1}{4(y-\sqrt{a-1})^2} + \frac{1}{4(y+\sqrt{a-1})^2} + \frac{1}{4 \sqrt{a-1} (y - \sqrt{a-1})} - \frac{1}{4 \sqrt{a-1} (y + \sqrt{a-1})} $$

so an antiderivative is

$$ -\frac{1}{4(y-\sqrt{a-1})} - \frac{1}{4(y+\sqrt{a-1})} + \frac{1}{4 \sqrt{a-1}} \ln \left(\frac{y-\sqrt{a-1}}{y+\sqrt{a-1}}\right)$$

Thus the integral is

$$ \frac{1}{4(\sqrt{a}-\sqrt{a-1})} + \frac{1}{4(\sqrt{a}+\sqrt{a-1})} - \frac{1}{4 \sqrt{a-1}} \ln \left(\frac{\sqrt{a}-\sqrt{a-1}}{\sqrt{a}+\sqrt{a-1}}\right)$$

which simplifies to

$$ \frac{\sqrt{a}}{2} - \frac{1}{2 \sqrt{a-1}} \ln(\sqrt{a} - \sqrt{a-1})$$

This is actually valid for all $a > 0$ except $a=1$, but for $a < 1$ you may want to use an alternative form to avoid complex numbers:

$$ \frac{\sqrt{a}}{2} + \frac{1}{2 \sqrt{1-a}} \arctan\left(\sqrt{1/a-1}\right)$$

For $a=1$ you can get the answer $1$ directly, or by taking the limit of either of these as $a \to 1$.

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be careful to treat the case $a=1$ and $a<1$ separatly – Jean-Sébastien Oct 16 '12 at 19:22
Thank you very much for taking your time to write it in details. – passerby51 Oct 17 '12 at 0:04

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