I don't know how to calculate this integral:

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

If we denote by $\Gamma$ a curve = semicircle centered at $0$ with radius $R$ + segment $[\ R, R]$, then the integral above is equal to $\lim_{R \rightarrow \infty} \int_{\Gamma} \frac{d z}{(1+z^2)^{n+1}}$ and the integral over the semicircle vanishes.

So I'm left with

$$\int_{\Gamma} \frac{d z}{(1+z^2)^{n+1}} = 2 \pi i \cdot res_i \frac{1}{(1 + z^2)^{n+1}} = 2 \pi i \frac{1}{(2i)^{n+1}} $$

What can I do now?

  • $\begingroup$ Are you interested in complex techniques only? $\endgroup$ – Mhenni Benghorbal Jan 24 '15 at 8:21
  • $\begingroup$ Yes, it's an exercise from complex analysis $\endgroup$ – Sasha Jan 24 '15 at 8:29

Keep in mind that, when $f(z)$ has an $(n+1)$th-order pole at $z=z_0$, the residue may be computed using

$$\operatorname*{Res}_{z=z_0} f(z) = \frac1{n!} \left [ \frac{d^n}{dz^n} \left [(z-z_0)^{n+1}f(z)\right ] \right ]_{z=z_0} $$

Thus in your case, as you are interested in the residue of the function $f(z) = (1+z^2)^{-(n+1)}$ at $z=i$, you must compute, as the integral,

$$\begin{align}\frac{i 2 \pi}{n!} \left [ \frac{d^n}{dz^n}(z+i)^{-(n+1)} \right ]_{z=i} &= \frac{i 2 \pi}{n!} (-1)^n\frac{(n+1)(n+2) \cdots (n+n-1) (n+n)}{(2 i)^{2 n+1}} \\ &= \pi \frac{(2 n)!}{2^{2 n} n!^2} \\ &= \frac{\pi}{2^{2 n}} \binom{2 n}{n} \end{align}$$

  • $\begingroup$ Of course, the pole has order $n+1$. Thanks a lot! $\endgroup$ – Sasha Jan 24 '15 at 9:23
  • $\begingroup$ Could you write down a general formula for $$\frac{d^n}{dz^n} \frac{1}{(z+i)^{n+1}}$$? $\endgroup$ – Sasha Jan 24 '15 at 9:40
  • $\begingroup$ @Sasha: Unfortunately, the formula I derived only works for the extremely special case where $(z-z_0)^{n+1} f(z)$ leaves you with a power of $z-z_1$ in the denominator. Otherwise, in most cases, a general formula is too difficult to derive this way. $\endgroup$ – Ron Gordon Jan 24 '15 at 9:45
  • $\begingroup$ I see, but how can I get from $$ \left [ \frac{d^n}{dz^n}(z+i)^{-(n+1)} \right ]_{z=i}$$ to $$(-1)^n\frac{(n+1)(n+2) \cdots (n+n-1) (n+n)}{(2 i)^{2 n+1}}$$? $\endgroup$ – Sasha Jan 24 '15 at 9:47
  • $\begingroup$ $$\frac{d}{dz} z^k = k z^{k-1} $$ $\endgroup$ – Ron Gordon Jan 24 '15 at 9:48

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