real analysis - $int_{0}^{infty} frac{1}{1 + x^r}:dx = frac{1}{r}Gammaleft( frac{r - 1}{r}right)Gammaleft( frac{1}{r}right)$

As part of a recent question I posted, I decided to try and generalise for a power of $2$ to any $r \in \mathbb{R}$. As part of the method I took, I had to solve the following integral:

$$\begin{equation} I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx \end{equation}$$

I believe what I've done is correct, but I'm concerned that I may missed something (in particular whether it holds for all $r \neq 0$). So, here I have two questions (1) Is what I've done correct? and (2) What other methods can be employed that doesn't rely on complex analysis?

Here is the method I took:

First make the substitution $u = x^{\frac{1}{r}}$ to arrive at

$$\begin{equation} I = \frac{1}{n} \int_{0}^{\infty} \frac{1}{1 + u} \cdot u^{1 -\frac{1}{r}}\:du \end{equation}$$

We now substitute $t = \frac{1}{1 + u}$ to arrive at:

$$\begin{align} I &= \frac{1}{r} \int_{1}^{0} t \cdot \left(\frac{1 - t}{t}\right)^{\frac{1}{r} -1}\frac{1}{t^2}\:dt = \frac{1}{r}\int_{0}^{1}t^{-\frac{1}{r}}\left(1 - t\right)^{ \frac{1}{r} - 1}\:dt \\ &= \frac{1}{r}B\left(1 - \frac{1}{n}, 1 + \frac{1}{r} - 1\right) = \frac{1}{r} B\left(\frac{r - 1}{r},\frac{1}{r}\right) \\ &= \frac{1}{r} B\left(\frac{r - 1}{r},\frac{1}{r}\right) \end{align}$$

Wheer $B(a,b)$ is the Beta function. Using the relationship between the Beta and Gamma function we arrive at:

$$\begin{equation} I = \frac{1}{r} \frac{\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right)}{\Gamma\left(\frac{r - 1}{r} + \frac{1}{r}\right)} = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right) \end{equation}$$

And so, we arrive at:

$$\begin{equation} I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right) \end{equation}$$

for $r > 1$

As per KemonoChen's comment and others, we can employ Euler's Reflection Formula to position this result for $\frac{1}{r} \not \in \mathbb{Z}$ Here, as $r \in \mathbb{R}, r > 1 \rightarrow \frac{1}{r} \not \in \mathbb{Z}$ and so our formula holds.

$$\begin{equation} I = \int_{0}^{\infty} \frac{1}{1 + x^r}\:dx = \frac{1}{r}\Gamma\left( \frac{r - 1}{r}\right)\Gamma\left( \frac{1}{r}\right) = \frac{\pi}{r\sin\left(\frac{\pi}{r} \right)} \end{equation}$$

Thank you also to Winther, Jjagmath, and MrTaurho's for their comments and corrections/clarifications.

Answer

Your given integral is closely related to the Mellin transform and can be evaluated by using Ramanujan's Master Theorem.

Ramanujan's Master Theorem

Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form
$$f(x)=\sum_{k=0}^{\infty}\frac{\phi(k)}{k!}(-x)^k$$ then the Mellin Transform of this function is given by
$$\int_0^{\infty}x^{s-1}f(x)dx=\Gamma(s)\phi(-s)$$

In order to get there we can expand the fraction as a geometric series

$$\begin{align} I=\int_0^{\infty}\frac1{1+x^n}dx&=\int_0^{\infty}\sum_{k=0}^{\infty}(-1)^k x^{kn}dx \end{align}$$

Now by applying the substitution $t=x^n$ followed by a little bit of reshaping yields to

$$\begin{align} I&=\int_0^{\infty}\sum_{k=0}^{\infty}(-1)^k x^{kn}dx\\ &=\frac1n\int_0^{\infty}t^{1/n-1}\sum_{k=0}^{\infty}(-1)^k\frac{k!}{k!}t^{k}dt\\ &=\frac1n\int_0^{\infty}t^{1/n-1}\sum_{k=0}^{\infty}(-1)^k\frac{\phi(k)}{k!}t^{k}dt \end{align}$$

Now we can apply Ramanujan's Master Theorem with $s=1/n$ and $\phi(k)=\Gamma(k+1)$ to get

$$\begin{align} I=\frac1n\int_0^{\infty}t^{1/n-1}\sum_{k=0}^{\infty}(-1)^k\frac{\phi(k)}{k!}t^{k}dt&=\frac1n\Gamma\left(\frac1n\right)\Gamma\left(1-\frac1n\right) \end{align}$$

And finally this can be simplified, as already pointed out, by using Euler's Reflection Formula $($as long as it holds i.e. for all $1/n\notin\mathbb Z$$)$ to get

$$I=\int_0^{\infty}\frac1{1+x^n}dx=\frac1n\frac{\pi}{\sin\left(\frac{\pi}{n}\right)}$$