TITLE: Seeking Methods to solve $\int_{0}^{\infty} \frac{e^{-x^n}}{x^n + 1}\:dx $
QUESTION [7 upvotes]: 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 $J_n(0) = \frac{1}{n}\Gamma\left(1 - \frac{1}{n}\right)\Gamma\left(\frac{1}{n}\right)$ (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 - 1\right)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}\Gamma\left(\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}\Gamma\left(\frac{1}{n}\right)
\end{equation}
Which yields the solution:
\begin{equation}
 J_n(t) = \frac{1}{n}\Gamma\left(1 - \frac{1}{n}, t\right)\Gamma\left(\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}\Gamma\left(1 - \frac{1}{n}, 1\right)\Gamma\left(\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.

REPLY [3 votes]: 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.