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Evaluate the integral

$$\int_{-\infty}^{\infty} \frac{1}{(1 + x^2)^{n+1}} dx. $$

I know that it equals $2\pi i$(the sum of the residues; at $z_k$) where $z_k$ are the poles of the function. I can evaluate this without the $n+1$ but that is throwing me for a serious loop.

Harry Peter
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jrmerrit
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3 Answers3

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For this case, consider

$$\oint_C \frac{dz}{(1+z^2)^{n+1}} $$

where $C$ is a semicircle of radius $R$ in the upper half plane. By the residue theorem, the contour integral is equal to $i 2 \pi$ time the residue at the pole $z=i$. Noting that the integral about the semicircle vanishes as $1/R^{2 n+1}$ as $R \to \infty$, we have

$$\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^{n+1}} = i 2 \pi \operatorname*{Res}_{z=i} \frac{1}{(1+z^2)^{n+1}}$$

Now,

$$\operatorname*{Res}_{z=i} \frac{1}{(1+z^2)^{n+1}} = \frac{1}{n!} \left [\frac{d^n}{dz^n} \frac{1}{(z+i)^{n+1}}\right ]_{z=i} = \frac{(2 n)!}{(n!)^2} \frac{(-1)^n}{(2 i)^{2 n+1}}$$

Therefore

$$\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^{n+1}} = \frac{\pi}{2^{2 n}} \binom{2 n}{n}$$

Ron Gordon
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  • You might want to add that the residue calculation involves the interpretation of $\frac{1}{2\pi\mathrm{i}}\oint_{(i)}\frac{\mathrm{d}z}{(z+\mathrm{i})^{n+1}(z-i)^{n+1}}$ as the series coefficient $c_n$ in the Taylor expansion $(z+\mathrm{i})^{-(n+1)}=\sum_{k=0}^\infty c_k (z-\mathrm{i})^k$. – ccorn Nov 21 '13 at 01:36
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    @ccorn: No, not really. I'm fine with what I posted. – Ron Gordon Nov 21 '13 at 01:43
  • +1 for this answer, as it takes the most direct path, which I feared would need to be harder than the answer I provided myself. – ccorn Nov 21 '13 at 17:03
  • @ccorn: thanks - I hope the +1 was because I convinced you that my way was good. In response to your input, I put a link to the residue result I used. – Ron Gordon Nov 21 '13 at 17:05
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A related technique. Here is an approach. Follow the steps

i)

$$ \int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^{n+1}}dx = 2\int_{0}^{\infty}\frac{dx}{(1+x^2)^{n+1}}dx $$

ii) Change of variables

$$ 1+x^2=\frac{1}{t} $$

iii) Use the $\beta$ function

$$ \mathrm{\beta}(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)},\quad \textrm{Re}(x), \textrm{Re}(y) > 0.\, $$

Community
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Mhenni Benghorbal
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For general $\Re n>-\frac{1}{2}$, see Mhenni Benghorbal's answer. However, if $n$ is a nonnegative integer, you can as well substitute $x=\cot\phi$, resulting in $$\int_{-\infty}^\infty \frac{1}{(1+x^2)^{n+1}}\mathrm{d}x = \int_0^\pi\sin^{2n}\phi\,\mathrm{d}\phi \stackrel{(*)}{=} \frac{1}{2}\int_0^{2\pi}\sin^{2n}\phi\,\mathrm{d}\phi$$ and then use this question and answer to calculate the last integral with the residue theorem. Note: the requirement that $n$ be an integer kicks in at (*).

Edit: I propose the above substitution because applying the residue theorem directly in the complex $x$-plane would require partial fraction decompositions depending on $n$ or a series expansion of $(1+x^2)^{-(n+1)}$, both of which is doable but requires more work.

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ccorn
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  • If you are speaking of direct evaluation of the residue in your edit, I assure you that it is very easy in this case. In case there is any doubt, this is exactly what I did in my solution. No partial fractions, no series expansions, just the $n$th derivative of a simple binomial. – Ron Gordon Nov 21 '13 at 12:11
  • @RonGordon: The $n$th derivative stems from Taylor's formula, hence from an implicit series expansion. You are right in that the underlying series expansion does not take additional calculations because only one of its coefficients is ultimately needed, and you have given that. I do like the brevity. But that series expansion is the theoretical justification without which your res$=$derivative "falls from heaven". I would have liked a few words of explanation. That's why I commented. Now I suspect that that sounded disrespectful. Sorry. – ccorn Nov 21 '13 at 16:37
  • Good one. Look, I don't care if you want to be blunt in assessing the deficiencies in my approach: I bloody well will do that to anyone else. I appreciate it. But what you wrote, both in your comment and the edit, made no sense to me. Now that I understand where you're coming from - I get it. If the OP asks, I will gladly supply the context. What I have posted, however, is a well-known, simple result that is central to residue theory. (http://en.wikipedia.org/wiki/Residue_(complex_analysis)#Limit_formula_for_higher_order_poles) It is in no way what I meant about "falling from heaven." – Ron Gordon Nov 21 '13 at 16:47
  • In that case, I did not feel the need to complicate or clutter my post with such details. if you look at even my longest posts, they are summaries, with details left out - or perhaps left to the end - because I value simplicity. Then the OP, or anyone else interested, can have a look and identify the piece they don't get. And perhaps it is that $n$th-derivative formula. I understand your point that it is a sort of leap - but a well-known one. Anyway...lots of math to do! Let's have fun doing it. – Ron Gordon Nov 21 '13 at 16:50