Detect the mistake: $\int_{-\infty}^{\infty}\frac{dx}{(x^{2}+4)^{n}}$

Question

Detect my mistake:
$\displaystyle\int_{-\infty}^{\infty}
\frac{{\rm d}x}{\left(x^{2} + 4\right)^{n}}\ $.

My try:

• We shall take a half circle around the upper part of the plane.
• The arc will tend to zero.
• Now we should calculate residue:
  $\displaystyle%
  \frac{\left(x – 2{\rm i}\right)^{n}}
  {\left(x^{2} + 4\right)^{n}} =
  \frac{1}{\left(x + 2{\rm i}\right)^{n}}$.
• We shall take $\left(n – 1\right)$ derivatives:
  $$
  \frac{\partial^{n – 1}}{\partial x^{n – 1}}
  \left[\frac{1}{\left(x + 2{\rm i}\right)^{n}}\right] =
  \frac{\left(2n – 2\right)!}{n!\left(x + 2{\rm i}\right)}
  = \frac{\left(2n – 2\right)!}{n!\left(4{\rm i}\right)}.
  $$
  Now the residue is incorrect, which leads to the integral result to be incorrect.

Does anyone see my mistake $?$.

Answer 1

$$I=\int_{-\infty}^\infty\frac{dx}{(x^2+4)^n}$$ $x=2u\Rightarrow dx=2du$ $$I=\int_{-\infty}^\infty\frac{2du}{(4u^2+4)^n}=2^{1-2n}\int_{-\infty}^\infty\frac{du}{(u^2+1)^n}=2^{2(1-n)}\int_0^\infty\frac{du}{(u^2+1)^n}$$ now try making the substitution $u=\tan t\Rightarrow du=\sec^2t\,dt$ so: $$I=2^{2(1-n)}\int_0^{\pi/2}\frac{\sec^2 t\,dt}{(\sec^2 t)^n}$$ now use the fact that: $$(\sec^2t)^{-n}=(\cos^2t)^n$$ so we have: $$I=2^{2(1-n)}\int_0^{\pi/2}\cos^{n-2}(t)dt$$ now you can use the beta function