Krdytkyu

As an extension of a question I posed earlier, I thought it would be best to try and final the result for a more general form:

begin{equation}
I_n = int_{0}^{infty} frac{e^{-x^n}}{x^n + 1}:dx
end{equation}
with $n in mathbb{R}, n > 1$

As with the previous question, I'm interested in finding alternative ways of solving this that does not rely on complex analysis.

My Method: I employ the exact same method as with my earlier question. Here first let

begin{equation}
J_n(t) = int_{0}^{infty} frac{e^{-tx^n}}{x^n + 1}:dx
end{equation}

We see that $I_n = J_n(1)$ and that $J_n(0) = frac{1}{n}Gammaleft(1 - frac{1}{n}right)Gammaleft(frac{1}{n}right)$ (This is shown here)

Now, take the derivative with respect to '$t$' to achieve
begin{align}
J_n'(t) &= int_{0}^{infty} frac{-x^ne^{-tx^n}}{x^n + 1}:dx = -int_{0}^{infty} frac{left(x^n + 1 - 1right)e^{-tx^n}}{x^n + 1}:dx \
&= -left[int_{0}^{infty}e^{-tx^n}:dx - int_{0}^{infty}frac{e^{-tx^n}}{x^n + 1}:dx right] \
&= -left[ frac{t^{-frac{1}{n}}}{n}Gammaleft(frac{1}{n}right) -J_n(t)right]
end{align}

Which yields the differential equation:

begin{equation}
J_n'(t) - J_n(t) = -frac{t^{-frac{1}{n}}}{n}Gammaleft(frac{1}{n}right)
end{equation}

Which yields the solution:

begin{equation}
J_n(t) = frac{1}{n}Gammaleft(1 - frac{1}{n}, tright)Gammaleft(frac{1}{n}right)e^t
end{equation}

And finally:

begin{equation}
I_n = J_n(1) = int_{0}^{infty} frac{e^{-x^n}}{x^n + 1}:dx = frac{e}{n}Gammaleft(1 - frac{1}{n}, 1right)Gammaleft(frac{1}{n}right)
end{equation}

Which for me, is a nice result. Fascinated to see other methods!

Edit - Thanks to spaceisdarkgreen for the pickup on my mistyping of the Incomplete Gamma Function.

I can offer a shorter way to obtain the final result in terms of the Incomplete Gamma Function. Therby consider the following representation of the Incomplete Gamma Function which can be found here on MSE

$$ Gamma(a,x)=frac{e^{-x}x^{a}}{Gamma(1-a)} int_0^infty frac{e^{-t} t^{-a}}{x+t} dttag1$$

Getting back to your original integral and applying the substitution $x^n=t$ yields to the following

$$begin{align}
I_n=int_{0}^{infty} frac{e^{-x^n}}{x^n + 1}dx&=int_{0}^{infty} frac{e^{-t}}{1+t}frac1nt^{1/n-1}dt\
&=frac1nint_0^{infty}frac{e^{-t}t^{-(1-1/n)}}{1+t}dt
end{align}$$

The latter integral is in the form of $(1)$ with $a=1-1/n$ and $x=1$ from where we can conclude that

$$begin{align}
I_n=frac1nint_0^{infty}frac{e^{-t}t^{-(1-1/n)}}{1+t}dt=frac1nfrac{Gammaleft(1-frac1n,1right)Gammaleft(frac1nright)}{e^{-1}}
end{align}$$

$$I_n=int_{0}^{infty} frac{e^{-x^n}}{x^n + 1}dx=frac enGammaleft(1-frac1n,1right)Gammaleft(frac1nright)$$

Of course this way of solving requires the knowledge of formula $(1)$ $($an impressive proof done by the user Felix Marin can be found within the linked post$)$ but nevertheless this evaluation method is quite compact.

I will offer up (again) a method that coverts the integral to a double integral. Note for convergence of the integral we require $n > 0$.

For $n > 0$, begin by enforcing a substitution of $x mapsto x^{1/n}$. This gives
$$I_n = frac{1}{n} int_0^infty frac{x^{1/n -1} e^{-x}}{1 + x} , dx qquad (1)$$

Noting that
$$frac{1}{x + 1} = int_0^infty e^{-u(x + 1)} , du,$$
the integral in (1) can be rewritten as
$$I_n = frac{1}{n} int_0^infty x^{1/n - 1} e^{-x} int_0^infty e^{-u(x + 1)} , du , dx,$$
or
$$I_n = frac{1}{n} int_0^infty e^{-u} int_0^infty x^{1/n - 1} e^{-x(u + 1)} , dx , du,$$
on changing the order of integration.

Enforcing a substitution of $x mapsto x/(u + 1)$ leads to
begin{align}
I_n &= frac{1}{n} int_0^infty (u + 1)^{-1/n} e^{-u} int_0^infty x^{1/n - 1} e^{-x} , dx , du\
&= frac{1}{n} Gamma left (frac{1}{n} right ) int_0^infty (u + 1)^{-1/n} e^{-u} , du.
end{align}
Finally, enforcing a substitution of $u mapsto u - 1$ one has
$$I_n = frac{e}{n} Gamma left (frac{1}{n} right ) int_1^infty u^{(1 - 1/n) - 1} e^{-u} , du = frac{e}{n} Gamma left (frac{1}{n} right ) Gamma left (1 - frac{1}{n}, 1 right ),$$
as expected.