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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.

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  • $\begingroup$ @J.G. - re reflection. I was hoping to have a result for all non zero real values. $\endgroup$ – user150203 Dec 16 '18 at 7:35
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    $\begingroup$ That doesn't look quite right to me... the first Gamma function should be an incomplete Gamma with lower bound $z=1.$ $\endgroup$ – spaceisdarkgreen Dec 16 '18 at 8:10
  • $\begingroup$ @spaceisdarkgreen - Thank you! yes yes, I mean't to type it as the incomplete beta at $z = 1$. Thanks for the pick up! $\endgroup$ – user150203 Dec 16 '18 at 8:12
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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

$$ \Gamma(a,x)=\frac{e^{-x}x^{a}}{\Gamma(1-a)} \int_0^\infty \frac{e^{-t} t^{-a}}{x+t} dt\tag1$$

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

$$\begin{align} I_n=\int_{0}^{\infty} \frac{e^{-x^n}}{x^n + 1}dx&=\int_{0}^{\infty} \frac{e^{-t}}{1+t}\frac1nt^{1/n-1}dt\\ &=\frac1n\int_0^{\infty}\frac{e^{-t}t^{-(1-1/n)}}{1+t}dt \end{align}$$

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

$$\begin{align} I_n=\frac1n\int_0^{\infty}\frac{e^{-t}t^{-(1-1/n)}}{1+t}dt=\frac1n\frac{\Gamma\left(1-\frac1n,1\right)\Gamma\left(\frac1n\right)}{e^{-1}} \end{align}$$

$$I_n=\int_{0}^{\infty} \frac{e^{-x^n}}{x^n + 1}dx=\frac en\Gamma\left(1-\frac1n,1\right)\Gamma\left(\frac1n\right)$$

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.

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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.

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  • $\begingroup$ Thank you so much for your solution. As discussed, I'm still developing my integral 'expansion' skills so it's great to see multiple examples. Please keep posting them up :-) $\endgroup$ – user150203 Dec 17 '18 at 4:25
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    $\begingroup$ Sure thing. Glad I can be of some help. Of course the conversion of an improper integral to a double integral is not always possible or appropriate, but like I said before, is a nice trick to have up one's sleeve. $\endgroup$ – omegadot Dec 17 '18 at 4:29

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