I am aware that $$\int_{0}^{\infty}e^{-x^2}\ dx = \frac{\sqrt{\pi}}{2}$$ but I was wondering if there was a general case for other exponents. Particularly: $$\int_{0}^{\infty}e^{-x^n} \ dx$$ where $n$ is a real number greater than $1$ (although I am not confident it would even work for non-integer values, so apologies in advance if that is the case).

Generally speaking, I have two questions concerning this integral:

$1.$ What is the new "solution" in terms of $n$? Can it be expressed in elementary terms (obviously not so for an indefinite integral, so this is assuming we use the previously stated bounds of $0$ and $\infty$).

$2.$ It would seem that as $n$ approaches $\infty$ the value of the indefinite integral approaches $1$. That is:

$$\lim_{n \to \infty} \int_{0}^{\infty}e^{-x^n} \ dx = 1$$ (Please correct me if I'm wrong). If this is so, and the integral is also equal to $1$ for case $n = 1$, for what value $n$ does the integral hold a minimum value (again, $1<n<\infty$)?

Thank you kindly!

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    $\begingroup$ $f_n$ defined on $[0,\infty)$ by $f_n(x) = e^{-x^n}$ converges pointwise to $f=\mathbb{1}_{[0,1)} + e^{-1}\mathbb{1}_{\{1\}}$. Now, from the dominated convergence theorem (the monotone convergence is probably enough; but the DCT is easy to apply) you get that $\int_0^\infty f_n \xrightarrow[n\to\infty]{} \int_0^\infty f = 1$. $\endgroup$ – Clement C. Jul 29 '16 at 17:44

The solution for $n>0$ is as follows: Let $x=u^{1/n}.$ Then $dx=\frac{1}{n}u^{1/n-1}\,du,$ so $$\int_0^\infty e^{-x^n}\,dx=\frac{1}{n}\int_0^\infty u^{1/n-1}e^{-u}\,du=\frac{1}{n}\Gamma(1/n)=\Gamma(1+1/n).$$ (See the Gamma function). Clearly as $n\to\infty$, the integral goes to $\Gamma(1)=1$.

The minimum of this function is difficult to nail down (due to the difficulty in differentiating the gamma function), but a numerical solution reveals that the minimum occurs at $n\approx 2.16623$ when the integral equals $0.8856$.

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    $\begingroup$ How do you get the first equality? $\endgroup$ – Gregory Grant Jul 29 '16 at 17:48
  • $\begingroup$ @Clement C. Sorry, I meant the minimum value. That was a foolish typo on my part. $\endgroup$ – Gizmo Jul 29 '16 at 17:50
  • $\begingroup$ @GregoryGrant See my edit for the appropriate substitution. $\endgroup$ – Alex S Jul 29 '16 at 17:53
  • $\begingroup$ Thanks I see it now, also can just subtitute $m=1+\frac1n$ into both sides to recover the usual definition of $\Gamma$. Neat. $\endgroup$ – Gregory Grant Jul 29 '16 at 17:54
  • $\begingroup$ @ClementC The integral you have written is, via a standard identity, equal to $\frac{1}{n}\Gamma(1/n)=\Gamma(1+1/n).$ $\endgroup$ – Alex S Jul 29 '16 at 17:54

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