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The real integral I am trying to compute with residues/contour integration is $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)^3} \,dx$ For $a$ positive and by using the complex integral

$$\int_{C_R}\frac{z^2}{(z^2+a^2)^3} \,dz$$

For some arc bounding only the positive pole, $a i$. I am not sure how to evaluate this pole of order 3.

In class we have discussed how we can write a function $f(z)$ with a pole of order $n$ at $z_0$ multiplicatively as $$f(z)=\frac{g(z)}{(z-z_0)^n}$$ or additively as $$f(z)=\frac{a_{-n}}{(z-z_0)^n}+\cdots+\frac{a_{-1}}{z-z_0}+h(z)$$

For holomorphic functions on a disc around the pole $h,g$, which I think is equivalent to finding a Laurent series

This yields the formula for the residue $$\text{Res}(f,z_0)=\lim_{z\rightarrow z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}[ (z-z_0)^nf(z)]$$

Which gets messy. Is there a more elegant way?

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  • $\begingroup$ Very sorry. I seemed to have made a serious mistake in my previous answer. I have just woken up but my brain perhaps hasn't. $\endgroup$ – Vim May 23 '16 at 0:21
  • $\begingroup$ You are right, my answer (now deleted lest it be misleading) shows only $$\frac1{(z\pm ai)^3}(\text{Const}+o(1))$$, so there is indeed nothing indicating any information about the residue. $\endgroup$ – Vim May 23 '16 at 0:23
  • $\begingroup$ @Vim no problem! Thank you for trying. $\endgroup$ – qbert May 23 '16 at 0:25
  • $\begingroup$ for integrating a rational function, the method that always works is the partial fraction decomposition, and it is really 99% equivalent to the residue method (the interesting part being seing how the two are related, when the contour is closed/closable) $\endgroup$ – reuns May 23 '16 at 2:41
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    $\begingroup$ I have reorganised my second answer that I uploaded a blurred photo in but eventually deleted nine hours ago. The gist there is using Taylor series instead of expansion, whose mathematical basis is the uniqueness of the Laurent series at the given point $ai$. Now I've undeleted it, and you can take a look if interested. $\endgroup$ – Vim May 23 '16 at 10:15
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Your calculation of the derivative is unnecessarily complicated. The reason we're multiplying by $(z-z_0)^n$ is that this is a factor in the denominator; so you can cancel it from the denominator:

\begin{align} \operatorname{Res}(f,z_0)&=\lim_{z\to z_0}\frac1{(n-1)!}\frac{\mathrm d^{n-1}}{\mathrm dz^{n-1}}\left[ (z-z_0)^nf(z)\right] \\ &=\lim_{z\to a\mathrm i}\frac12\frac{\mathrm d^2}{\mathrm dz^2}\left[ (z-a\mathrm i)^3\frac{z^2}{\left(z^2+a^2\right)^3}\right] \\ &= \lim_{z\to a\mathrm i}\frac12\frac{\mathrm d^2}{\mathrm dz^2}\frac{z^2}{(z+a\mathrm i)^3} \\ &=\lim_{z\to a\mathrm i}\frac12\left(\frac2{(z+a\mathrm i)^3}-2\cdot\frac{6z}{(z+a\mathrm i)^4}+\frac{12z^2}{(z+a\mathrm i)^5}\right) \\ &=\frac{\mathrm i}{2a^3}\left(\frac14-\frac34+\frac38\right) \\ &=-\frac{\mathrm i}{16a^3}\;. \end{align}

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  • $\begingroup$ I had mathematica do it :P. I missed that I could simplify as you did from step 2 to 3. I do find it odd that my computation yielded zero as the residue though $\endgroup$ – qbert May 23 '16 at 1:52
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@joriki posted the same answer I was going to before I could get it typed in, so I'll just post the second half: $$\begin{align}\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)^3}dx&=\frac2{a^3}\int_0^{\infty}\frac{y^2}{(y^2+1)^2}dy=\frac1{a^3}\int_0^{\infty}\frac{z^{\frac12}}{(z+1)^3}dz\\ &=\frac1{a^3}\int_0^1(1-t)^{\frac12}t^{\frac12}dt=\frac1{a^3}\text{B}\left(\frac32,\frac32\right)\\ &=\frac1{a^3}\frac{\Gamma\left(\frac32\right)\Gamma\left(\frac32\right)}{\Gamma(3)}=\frac1{a^3}\frac{\left(\frac12\pi\right)\left(\frac12\pi\right)}{2}=\frac{\pi}{8a^3}\end{align}$$ In the above we have made the successive substitutions $x=ay$, $y^2=z$, and $t=1/(z+1)$ and made use of the symmetry of the original integral. Then I made use of the properties of the Beta and Gamma function.
So this provides a useful check of the contour integration results.

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  • $\begingroup$ very cool use of the gamma and beta functions, I have been meaning to check them out, I will post my answer using more elementary methods in a minute $\endgroup$ – qbert May 23 '16 at 2:39
  • $\begingroup$ Actually Beta functions aren't required in this case because the substitution $t=\sin^2\theta$ puts the integral in a form where a primitive is in sight. But from the start the chain of substitutions was designed to get to a Beta function, so I didn't think of anything else before posting. $\endgroup$ – user5713492 May 23 '16 at 3:21
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After two failed attempts, and since it looks like no one has proposed Taylor series method, which is my favourite, I can't resist posting this not-so-elegant answer.

Computationally this answer might not be the most economical one, but I think in general it should be less painful than differentiating twice or even more. So here it goes \begin{align} f(z) &=\frac{z^2}{(z^2+a^2)^3} = \frac{1}{(z-ai)^3}\frac{(z-ai+ai)^2}{(z-ai+2ai)^3} \\ & = \frac{1}{(z-ai)^3}\frac{(ai)^2}{(2ai)^3}\frac{(1+\frac{z-ai}{ai})^2}{(1+\frac{z-ai}{2ai})^3} \\ & = \frac{1}{(z-ai)^3}\frac1{8ai}(1+2\frac{z-ai}{ai}+\frac{(z-ai)^2}{(ai)^2})(1-3\frac{z-ai}{2ai}+6\frac{(z-ai)^2}{(2ai)^2}+O(|z-ai|^3)).\\ \end{align} Note that in the last equality we have used the Newton formula $$(1+w)^{-3}=1-3w+6w^2+O(|w|^3),$$ for small $w$.

Now, we see that $f(z)$ is $\frac{1}{(z-ai)^3}$ times lots of other stuff. But all we want is only the residue, so anyway amid all that stuff all we need are terms of magnitude of $(z-ai)^2$ (which is also why we only need to expand up to $O(|w|^3)$ in the Newton formula). Pick out all the needed terms from that stuff (which is essentially a very simple combinatoric task), we get \begin{align} f(z) & = \frac{1}{(z-ai)^3} \frac1{8ai}(6\frac1{(2ai)^2}-2\cdot 3\frac{1}{2(ai)^2}+\frac{1}{(ai)^2}+\cdots) \\ & = \frac{1}{(z-ai)^3} \frac1{8(ai)^3} (-\frac12(z-ai)^2+\cdots), \end{align} which yields our desired results as $$\text{Res}(f,ai)=\frac1{16a^3 i}.$$

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Computing the residue as @joriki suggested \begin{align*} \text{Res}(f,z_0)&=\lim_{z\rightarrow z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}[ (z-z_0)^nf(z)]\\ &\Rightarrow \text{Res}(f,ai)=\frac{1}{2}\lim_{z\rightarrow ai}\frac{d^2}{dz^2} [(z-ai)^{3}\frac{z^2}{(z^2+a^2)^3}]= \frac{1}{2}\lim_{z\rightarrow ai}\frac{d^2}{dz^2} [(z-ai)^{3}\frac{z^2}{(z-ai)^3(z+ai)^3}]\\ &=\frac{1}{2}\lim_{z\rightarrow ai}\frac{d^2}{dz^2} [\frac{z^2}{(z+ai)^3}]=\frac{1}{2}\lim_{z\rightarrow ai}\frac{d}{dz} [\frac{z(2ai-z)}{(z+ai)^4}]\\ &=\frac{1}{2}\lim_{z\rightarrow ai}\frac{2(ai-z)}{(z+ai)^4} -4\frac{z(2ai-z)}{(z+ai)^5}= -4\frac{1}{2}\lim_{z\rightarrow ai}\frac{z(2ai-z)}{(z+ai)^5}\\ &=-2\frac{(ai)^2}{(2ai)^5}=\frac{1}{16ia^3} \end{align*} yielding then for our integral: \begin{align*} \int_{C_R}\frac{z^2}{(z^2+a^2)^3}dz&=2\pi i \text{Res}(f,ai)=\frac{2\pi i}{16ia^3} =\frac{\pi}{8a^3} \end{align*} Now, to send the arc legnth to zero, we parametrize the arc as usual $\gamma_R(t)=Re^{it}$, $t\in [0,\pi]$, and bounding our integrand using the max and path length \begin{align*} |\int_{\gamma_R}\frac{z^2}{(z^2+a^2)^3}&|\leq \max_{|z|=R}|\frac{z^2}{(z^2+a^2)^3}|*||\gamma_R(t)||= \frac{R^2}{(R^2+a^2)^3}\pi R\sim cR^{-3}\rightarrow 0\;\text{as}\; R\rightarrow \infty \end{align*} Yielding our answer: \begin{align*} \lim_{R\rightarrow \infty}\int_{-R}^{R}\frac{z^2}{(z^2+a^2)^3}dz=\int_{-\infty}^{\infty}\frac{z^2}{(z^2+a^2)^3}dz=\frac{\pi}{8a^3} \end{align*}

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