Question

For positive integers n, p and q, define

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

Answer 1

$\frac{\sqrt{\pi} \Gamma(n/2)}{2^{n+1} \Gamma((n+1)/2)}$

Answer 2

The integral is given by $I_n = \int_{0}^{\infty} \frac{1}{(e^x + e^{-x})^n} dx$. We know that $e^x + e^{-x} = 2 \cosh(x)$. So, $I_n = \int_{0}^{\infty} \frac{1}{(2 \cosh(x))^n} dx = \frac{1}{2^n} \int_{0}^{\infty} \frac{1}{\cosh^n(x)} dx = \frac{1}{2^n} \int_{0}^{\infty} \text{sech}^n(x) dx$.

Let's evaluate the integral $J_n = \int_{0}^{\infty} \text{sech}^n(x) dx$. Use the substitution $u = \tanh(x)$. Then $du = \text{sech}^2(x) dx$. Also, $\text{sech}^2(x) = 1 - \tanh^2(x) = 1 - u^2$. The limits of integration change from $x=0$ to $u=\tanh(0)=0$ and from $x \to \infty$ to $u=\tanh(\infty)=1$. $J_n = \int_{0}^{\infty} \text{sech}^{n-2}(x) \text{sech}^2(x) dx = \int_{0}^{1} (1 - \tanh^2(x))^{(n-2)/2} du = \int_{0}^{1} (1 - u^2)^{(n-2)/2} du$.

Now, let's use the substitution $u = \sin \theta$. Then $du = \cos \theta d\theta$. The limits of integration change from $u=0$ to $\sin \theta = 0 \implies \theta = 0$ and from $u=1$ to $\sin \theta = 1 \implies \theta = \pi/2$. The integral becomes: $\int_{0}^{\pi/2} (1 - \sin^2 \theta)^{(n-2)/2} \cos \theta d\theta = \int_{0}^{\pi/2} (\cos^2 \theta)^{(n-2)/2} \cos \theta d\theta = \int_{0}^{\pi/2} \cos^{n-2} \theta \cos \theta d\theta = \int_{0}^{\pi/2} \cos^{n-1} \theta d\theta$.

This is a standard integral, a form of Wallis' integral. Let $K_m = \int_{0}^{\pi/2} \cos^m \theta d\theta$. $K_m = \begin{cases} \frac{m-1}{m} \frac{m-3}{m-2} \cdots \frac{1}{2} \frac{\pi}{2} & \text{if m is even} \\ \frac{m-1}{m} \frac{m-3}{m-2} \cdots \frac{2}{3} & \text{if m is odd} \end{cases}$. In our case, $m = n-1$.

So, $J_n = \int_{0}^{\pi/2} \cos^{n-1} \theta d\theta$.

Case 1: $n$ is even. Let $n = 2k$ for some positive integer $k$. Then $m = n-1 = 2k-1$ (odd). $J_{2k} = \int_{0}^{\pi/2} \cos^{2k-1} \theta d\theta = \frac{2k-2}{2k-1} \frac{2k-4}{2k-3} \cdots \frac{2}{3}$. This can be written using double factorials: $J_{2k} = \frac{(2k-2)!!}{(2k-1)!!}$. Then $I_{2k} = \frac{1}{2^{2k}} J_{2k} = \frac{1}{2^{2k}} \frac{(2k-2)!!}{(2k-1)!!}$.

Case 2: $n$ is odd. Let $n = 2k+1$ for some non-negative integer $k$. Then $m = n-1 = 2k$ (even). $J_{2k+1} = \int_{0}^{\pi/2} \cos^{2k} \theta d\theta = \frac{2k-1}{2k} \frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2}$. This can be written using double factorials: $J_{2k+1} = \frac{(2k-1)!!}{(2k)!!} \frac{\pi}{2} = \frac{(2k-1)!!}{2^k k!} \frac{\pi}{2}$. Then $I_{2k+1} = \frac{1}{2^{2k+1}} J_{2k+1} = \frac{1}{2^{2k+1}} \frac{(2k-1)!!}{(2k)!!} \frac{\pi}{2} = \frac{(2k-1)!!}{2^{2k+1} 2^k k!} \frac{\pi}{2} = \frac{(2k-1)!!}{2^{3k+1} k!} \frac{\pi}{2}$. A more common way to express this is using Gamma functions, which covers both even and odd cases. $J_n = \int_{0}^{\pi/2} \cos^{n-1} \theta d\theta$. Using the relation between Wallis' integral and the Gamma function: $\int_{0}^{\pi/2} \sin^p x \cos^q x dx = \frac{\Gamma(\frac{p+1}{2}) \Gamma(\frac{q+1}{2})}{2 \Gamma(\frac{p+q+2}{2})}$. Here, $p=0$ and $q=n-1$. $J_n = \frac{\Gamma(\frac{0+1}{2}) \Gamma(\frac{(n-1)+1}{2})}{2 \Gamma(\frac{0+(n-1)+2}{2})} = \frac{\Gamma(1/2) \Gamma(n/2)}{2 \Gamma((n+1)/2)}$. Since $\Gamma(1/2) = \sqrt{\pi}$, $J_n = \frac{\sqrt{\pi} \Gamma(n/2)}{2 \Gamma((n+1)/2)}$.

Finally, $I_n = \frac{1}{2^n} J_n = \frac{1}{2^n} \frac{\sqrt{\pi} \Gamma(n/2)}{2 \Gamma((n+1)/2)} = \frac{\sqrt{\pi} \Gamma(n/2)}{2^{n+1} \Gamma((n+1)/2)}$.

This formula gives the value of $I_n$ for any positive integer $n$.