Csdrhrt

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 - tright)^{ frac{1}{r} - 1}:dt \
&= frac{1}{r}Bleft(1 - frac{1}{n}, 1 + frac{1}{r} - 1right) = frac{1}{r} Bleft(frac{r - 1}{r},frac{1}{r}right) \
&= frac{1}{r} Bleft(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{Gammaleft( frac{r - 1}{r}right)Gammaleft( frac{1}{r}right)}{Gammaleft(frac{r - 1}{r} + frac{1}{r}right)} = frac{1}{r}Gammaleft( frac{r - 1}{r}right)Gammaleft( 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}Gammaleft( frac{r - 1}{r}right)Gammaleft( 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}Gammaleft( frac{r - 1}{r}right)Gammaleft( frac{1}{r}right) = frac{pi}{rsinleft(frac{pi}{r} right)}
end{equation}

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

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

$$begin{align}
I&=int_0^{infty}sum_{k=0}^{infty}(-1)^k x^{kn}dx\
&=frac1nint_0^{infty}t^{1/n-1}sum_{k=0}^{infty}(-1)^kfrac{k!}{k!}t^{k}dt\
&=frac1nint_0^{infty}t^{1/n-1}sum_{k=0}^{infty}(-1)^kfrac{phi(k)}{k!}t^{k}dt
end{align}$$

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

$$begin{align}
I=frac1nint_0^{infty}t^{1/n-1}sum_{k=0}^{infty}(-1)^kfrac{phi(k)}{k!}t^{k}dt&=frac1nGammaleft(frac1nright)Gammaleft(1-frac1nright)
end{align}$$

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/nnotinmathbb Z$$)$ to get

$$I=int_0^{infty}frac1{1+x^n}dx=frac1nfrac{pi}{sinleft(frac{pi}{n}right)}$$

Once again I will offer up a method that first converts the integral to a double integral.

For $r > 0$, we begin by enforcing a substitution of $x mapsto x^{1/r}$. Doing so yields
$$I = frac{1}{r} int_0^infty frac{x^{1/r - 1}}{1 + x} , dx.$$

Now noting that
$$frac{1}{1 + x} = int_0^infty e^{-u(1 + x)} , du,$$
our integral can be rewritten as
$$I = frac{1}{r} int_0^infty x^{1/r - 1} int_0^infty e^{-u (1 + x)} , du , dx,$$
or
$$I = frac{1}{r} int_0^infty e^{-u} int_0^infty x^{1/r - 1} e^{-ux} , dx , du,$$
after changing the order of integration.

Next we enforce a substitution of $x mapsto x/u$. This gives
begin{align}
I &= frac{1}{r} int_0^infty u^{- 1/r} e^{-u} , du int_0^infty x^{1/r - 1} e^{-x} , dx\
&= frac{1}{r} Gamma left (1 - frac{1}{r} right ) Gamma left (frac{1}{r} right )\
&= frac{pi}{r sin left (frac{pi}{r} right )},
end{align}
where in the last line we have made use of Euler's reflexion formula for the gamma function.

NOT A FULL SOLUTION:

I've been working with special cases of the integral.

Here we will consider $r = 2m$ where $m in mathbb{N}$. In doing so, we observe that the roots of the denominator are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:

begin{align}
x^{2m} + 1 = 0 rightarrow x^{2m} = e^{pi i}
end{align}

By De Moivre's formula, we observe that:

begin{align}
x = expleft({frac{pi + 2pi j}{2m} i} right) mbox{ for } j = 0dots 2m - 1
end{align}

Which we can express as the set

begin{align}
S &= Bigg{ expleft({frac{pi + 2pi cdot 0}{2m} i} right) , :expleft({frac{pi + 2pi cdot 1}{2m} i} right),dots,:expleft({frac{pi + 2pi cdot (2m - 2)}{2m} i} right)\
&qquad:expleft({frac{pi + 2pi cdot (2m - 1)}{2m} i} right)Bigg} \
end{align}

Which can be expressed as the set of 2-tuples

begin{align}
S &= left{ left( expleft({frac{pi + 2pi j}{2m} i} right) , :expleft({frac{pi + 2pi(2m - 1 - j )}{2m} i} right)right): bigg|: j = 0 dots m - 1right}\
& = left{ (z_j, cleft(z_jright):|: j = 0 dots m - 1 right}
end{align}

From here, we can factor $x^{2m} + 1$ into the form

begin{align}
x^{2m} + 1 &= prod_{r in S} left(x + r_jright)left(x + c(r_j)right) \
&= prod_{r in S} left(x^2 + left(r_j + c(r_j)right)x + r_j c(r_j)right) \
&= prod_{r in S} left(x^2 + 2Releft(r_jright)x + left|r_j right|^2right)
end{align}

For our case here $left|r_j right|^2 = 1$ and $Releft(r_jright) = cosleft({frac{pi + 2pi j}{2m} } right)$. Hence,

begin{align}
frac{1}{x^{2m} + 1} = prod_{j = 0}^{m - 1}frac{1}{ x^2 + 2cosleft({frac{pi + 2pi j}{2m} } right)x + 1}
end{align}

From here, to evaluate the integral we must employ Partial Fraction Decomposition:

begin{align}
frac{1}{x^{2m} + 1} = prod_{j = 0}^{m - 1}frac{1}{ x^2 + 2cosleft({frac{pi + 2pi j}{2m} } right)x + 1} = sum_{j = 0}^{m - 1}frac{alpha_j + beta_jx}{ x^2 + 2cosleft({frac{pi + 2pi j}{2m} } right)x + 1}
end{align}

And solve for $alpha_j$ and $beta_j$. Putting the coefficents to the side we can find general expressions for the integral:

begin{align}
frac{1}{x^{2m} + 1} = prod_{j = 0}^{m - 1}frac{1}{ x^2 + 2cosleft({frac{pi + 2pi j}{2m} } right)x + 1}
end{align}

From here, to evaluate the integral we must employ Partial Fraction Decomposition:

begin{align}
int_{0}^{infty}frac{1}{x^{2m} + 1}:dx &= int_{0}^{infty}sum_{j = 0}^{m - 1}frac{alpha_j + beta_jx}{ x^2 + 2cosleft({frac{pi + 2pi j}{2m} } right)x + 1}:dx \
&= sum_{j = 0}^{m - 1}left[ int_{0}^{infty}frac{alpha_j}{ x^2 + 2cosleft({frac{pi + 2pi j}{2m} } right)x + 1}:dx + int_{0}^{infty}frac{beta_jx}{ x^2 + 2cosleft({frac{pi + 2pi j}{2m} } right)x + 1}:dxright] \
&=sum_{j = 0}^{m - 1}left[ I_1 + I_2right]
end{align}

Evaluating each individually:
begin{align}
int_{0}^{infty} frac{alpha_j}{ x^2 + 2cosleft(frac{pi + 2pi j}{2m} right)x + 1}:dx &= left[ cscleft(frac{pi + 2pi j}{2m} right)arctanleft(frac{(x - 1)tanleft(frac{pi + 2pi j}{4m} right)}{x + 1} right)right]_{0}^{infty} \
&= cscleft(frac{pi + 2pi j}{2m} right)left( frac{pi + 2pi j}{2m}right)
end{align}

Now if we consider the second integral on it's own we find that as a general expression the integral doesn't converge. This doesn't take away from this method, but to proceed we must solve for the unknown coefficients to proceed.