4

I want to compute $$ \int_{-\infty}^\infty \frac 1{ (1+x^2)^{n+1}} dx $$ for $n \in \mathbb N_{\geq 1}$. If I let $$ f(z) := \frac 1 {(z+i)^{n+1}(z-i)^{n+1}} $$ then I see that $f$ has poles of order $n+1$ in $\pm i$. Initially I thought that $$ [-R,R] \cup \{R \exp(i \theta) \mid \theta \in [0,\pi] \} $$ would be a good conture but for the residue in $i$ I get $(-1)^n (n+1) 2 i$. I know that the result must be $$ \frac {1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdots 2 n} \pi $$ which looks quite different than my residue. What goes wrong here ?

Seirios
  • 33,157

4 Answers4

2

I think that both of you had made mistake in calculating derivatives. $$\underset{z=i}{\operatorname{res}}f(z)=\underset{z=i}{\operatorname{res}}\frac{1}{\left(z+i\right)^{n+1}\left(z-i\right)^{n+1}}=\frac{1}{n!}\underset{z\to i}\lim \frac{d^n}{dz^n}\left(\frac{\left(z-i\right)^{n+1}}{\left(z-i\right)^{n+1}\left(z+i\right)^{n+1}}\right)=\\ =\frac{1}{n!}\underset{z\to i}\lim \frac{d^n}{dz^n}\left(\frac{1}{\left(z+i\right)^{n+1}}\right)=\frac{1}{n!}\underset{z\to i}\lim \frac{d^{n-1}}{dz^{n-1}}\left(\frac{(-1)(n+1)}{\left(z+i\right)^{n+2}}\right)=\\ =\frac{1}{n!}\underset{z\to i}\lim \frac{d^{n-2}}{dz^{n-2}}\left(\frac{\left(-1\right)^2(n+1)(n+2)}{\left(z+i\right)^{n+3}}\right)=...=\frac{1}{n!}\underset{z\to i}\lim \left(\frac{\left(-1\right)^n(n+1)(n+2)..2n}{\left(z+i\right)^{2n+1}}\right)=\\ =\frac{1}{n!}\left(\frac{\left(-1\right)^n(n+1)(n+2)..2n}{2^{2n+1}i^{2n+1}}\right)$$

cool
  • 1,089
  • Is this times $2 \pi i$ (the real part of it) equal to what I wrote that it should be ?. Indeed your residue looks correct to me. –  Jun 30 '13 at 12:57
  • When you evaluate this residues up to the end (one more), add them and multiply the summ with $2\pi i$ (as it is said in the theorem Cauchy of residues), you'll get your correct answer. – cool Jun 30 '13 at 13:11
  • Why them ? Isn't there only one residue to calculate ? –  Jun 30 '13 at 13:19
  • Yes, you are right, only one. Now that was my mistake. – cool Jun 30 '13 at 13:28
  • Ok thanks. I will try to work it out :) –  Jun 30 '13 at 13:29
  • It works. $2 \pi i Res_{z = i} f(z) = \pi \frac {(2n)!}{(n!)^2 2^{2n}} $ but this is $$ \frac {(2n)!}{(2 \cdot 4 \cdot 6 \cdots \cdot 2n)^2} = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdots 2n} $$ –  Jun 30 '13 at 15:58
  • @Paul: it is hard to see, but if you multiply your residue by $2\pi i$, you get the same answer as everyone else. – robjohn Jun 30 '13 at 17:27
2

To compute the residue at $z=i$ of $$ \begin{align} \frac1{(1+z^2)^{n+1}} &=\frac1{(z+i)^{n+1}}\frac1{(z-i)^{n+1}}\\ &=\frac1{((z-i)+2i)^{n+1}}\frac1{(z-i)^{n+1}} \end{align} $$ we need to compute the coefficient of $(z-i)^n$ in $\frac1{((z-i)+2i)^{n+1}}$.

Using the binomial theorem, the full term is $$ \begin{align} \binom{-n-1}{n}\left(\frac{z-i}{2i}\right)^n(2i)^{-n-1} &=(z-i)^n(-1)^n\binom{2n}{n}(2i)^{-2n-1}\\ &=(z-i)^n\color{#C00000}{\frac1i\binom{2n}{n}2^{-2n-1}} \end{align} $$ Thus, the integral along the contour which follows the real axis and then circles back around through the upper half-plane is $2\pi i$ times $\color{#C00000}{\text{this residue}}$. That is, $$ \int_{-\infty}^\infty\frac{\mathrm{d}x}{(1+x^2)^{n+1}}=\frac\pi{4^n}\binom{2n}{n} $$

robjohn
  • 345,667
1

Here is another approach to calculating the residue of $f(z) = (1+z^2)^{-n-1}$ at $z = i $ which uses the Binomial Theorem to avoid taking derivatives. It will simplify notation a bit if we let $w = z - i$ and calculate the residue of $f(w)$ at $w = 0$. So consider $$f(w) = [1 + (w+i)^2]^{-n-1} = w^{-n-1} (2i)^{-n-1} \left(1 + \frac{w}{2i} \right) ^{-n-1}$$ $$= w^{-n-1} (2i)^{-n-1} \sum_{k=0}^{\infty} \binom{-n-1}{k} (2i)^{-k} w^k $$ The coefficient of $w^{-1}$ corresponds to $k=n$ in the summation, so $$Res_{w=0} f(w)= (2i)^{-n-1} (2i)^{-n} \binom{-n-1}{n} = -2^{-2n-1} (-1)^n \binom{-n-1}{n} i= -2^{-2n-1} \binom{2n}{n} i$$

awkward
  • 14,736
  • Awesome. Great thanks for that really nice approach. I love it to work with just a power series :) –  Jun 30 '13 at 16:20
  • Ah, I just noticed that my answer is essentially the same as yours (+1). – robjohn Jun 30 '13 at 17:43
0

I believe your residue is wrong.

If $f$ has a pole of order $N$ in $z=a$, so that $f(z)=(z-a)^{-N}g(z)$ (where $g$ is a function which is analytic in a neighbourhood of $z=a$), then the residue of $f$ in $z=a$ is $$\operatorname{Res}\limits_{z=a} f(z) = \frac{g^{(N-1)}(a)}{(N-1)!}.$$

In your case, $$g^{k}(z)=(-1)^k(n+1)(n+2)\cdots(n+k)(z+i)^{-(n+k+1)},$$ so that $$\operatorname{Res}\limits_{z=i} f(z) = \frac{(-1)^n(n+1)(n+2)\cdots2n}{n!(2i)^{2n+1}}=\frac{(2n)!}{n!n!2^{2n}2i}=\frac{1}{2i2^{2n}}\binom{2n}{n}.$$ Thus, the integral becomes $$\frac{\pi}{2^{2n}}\binom{2n}{n}.$$

Mårten W
  • 3,480