hmm , can i show you an exercise im solving ? i keep getting someting close to the answer , not sure why. (only if its ok with you.. i dont want to bother you)
i defined $f(z) = 1/(z\ ^ 2 +1) \ ^ {n+1}$ and the integral above is equal to $2\pi i Res(f,i)$. $Res(f,i) = \dfrac{1}{n!} lim_{z\to i}\dfrac{d \ ^ n }{d z \ ^ n}[(z-i) \ ^ {n+1} f(z)] $
(i did the usual trick with the half circle )
Now what im geting is that $Res(f,i) = \dfrac{(n+1)(n+2) \dots (2n) (-1)\ ^ n (2i) \ ^ {-2n-1}}{n!}$
this do not look like the RHS in the picture ^^
(even after i multiply it by $2\pi i $)
what im doing wrong? i went over it again and again.. nothing seems wrong to me
On the one hand, it's usually easier to find the residue via the Laurent expansion. On the other hand, this problem offers itself to induction. Start with $n = 0$ and integrate by parts to obtain a recurrence. But as for your result, cancel the $(-1)^n$ from the numerator against the $i^{2n}$ from the denominator. Multiply numerator and denominator each with $n!$. You get $\binom{2n}{n}$. In the denominator, you have a $2^{2n}$ left, pair one $2$ with each factor of the two $n!$,
that gives $(2\cdot 4 \cdot \dotsc \cdot (2n))\cdot (2\cdot 4 \cdot \dotsc \cdot (2n))$. Cancel one of these against the even factors of $(2n)!$ in the numerator. Multiply with $2\pi i$, and ta-daa!
