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I was wondering what the best way to solve this would be. I was using the cauchy integral residue formula which revolved around taking 2 derivatives since the poles are at +/- i and are of order 3. Quickly realized that taking double derivatives are pretty cumbersome. Is there a better way to solve it?

0<a=1

$\int_{-\infty}^\infty \frac{(z^{a-1})}{(z^2+1)^3}dz$

$I=\int_{C}\frac{(z^{a-1})}{(z^2+1)^3}dz$

Cauchy integral residue integral formula

qwerty2277
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  • Welcome to math.SE. Please see here about relying on images for essential parts of your question. – joriki Mar 26 '23 at 06:50
  • There is but please note you cannot directly apply the contour integration if $a$ is not an integer since the integrand won’t be meromorphic – FShrike Mar 26 '23 at 07:35
  • @FShrike, forgot to include that 0<a<=1 which I assume that a doesn't have to be an integer – qwerty2277 Mar 26 '23 at 07:38

1 Answers1

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First make the substitution $z= e^{2u}$ to cast the integrand into the form $\frac{ e^{\beta u} \ du}{ \cosh^3 u}$ where $\beta$ is real.

Then note that in terms of the complex variable $w=u+iv$
$\cosh w = \cos (iw)$ is periodic with period $2\pi$ in $v$. This allows you to create a closed loop contour in the $w$ plane comprising two parallel horizontal lines. The loop encloses one singularity at $w= i\pi/2$. Thus you are reduced to computing a single residue, at that triple pole.

To find the residue at the triple pole you need to find a Laurent expansion. You can exploit the trig identity $\cos (\pi/2+ \zeta)= -\sin \zeta =- \zeta+ \ldots$ and cube this to determine the Laurent expansion of the integrand there.

Note that the two vertical segments in your contour are harmless if the denominator grows much faster than the numerator. This will be true for a restricted range of $\beta $ values.

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