Integration of $\int_0^\infty \sqrt x e^{-x^2} dx$

Question

I am having a really hard time trying to figure out what to do with this. I feel like I've tried everything but I'm obviously missing something. Any suggestions?

$$\int_0^\infty \sqrt x\ e^{-x^2}\ dx$$

These are all the techniques I have learned in class/can choose from: integration by parts, u-substitution, partial fractions, trig substitution. I don't know anything about the gamma function and am not allowed to use it.

Try a substitution of $u=\sqrt x$. Realize a gamma integral once you have done that. — G.Dinesh Nathan, Jun 06 '14 at 06:07
@G.DineshNathan check the question, he specifically says he cannot use the gamma function. — DanZimm, Jun 06 '14 at 06:08
see http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf for new tricks... — James S. Cook, Jun 06 '14 at 06:09
@G.DineshNathan Also, you probably meant $u = x^2$, as $u = \sqrt{x}$ would make the exponent $u^4$. — M. Vinay, Jun 06 '14 at 06:10
I doubt there is a way to solve this without the Gamma function. — Pranav Arora, Jun 06 '14 at 06:16
The answer is $\Gamma(3/4)/2$, and this can't be expressed using "elementary" functions. So if he can't use the Gamma function he's out of luck. Unless perhaps the question was whether this converges, rather than find the value? — Robert Israel, Jun 06 '14 at 06:17
@RobertIsrael Or perhaps he's expected to express the answer as a series? — David H, Jun 06 '14 at 06:23
If you have learned about infinite series, you an express this as such a series and integrate term by term. Other than that, there is no way that I know of to evaluate this integral. — A. Thomas Yerger, Jun 06 '14 at 06:43
You can find a series for $e^{-x^{2}}$ and multiply through. — A. Thomas Yerger, Jun 06 '14 at 16:36
If you mean a Taylor series, the terms will not have finite integrals from $0$ to $\infty$. — Robert Israel, Jun 06 '14 at 21:23
Maybe he was actually given $\int_0^\infty\sqrt x e^{-x^3},dx$? That one can be converted into a Gaussian integral, and this site features many clever, elementary evaluations of it. — Antonio DJC, Mar 07 '18 at 07:37

score 2 · Answer 1 · answered Jun 06 '14 at 13:16

Let $~t=x^{^\tfrac32}.~$ Then $\displaystyle\int_0^\infty\sqrt x~e^{-x^2}~dx=\frac23\int_0^\infty\exp\bigg(-t^{^\tfrac43}\bigg)~dt=\frac23\cdot\bigg(\frac34\bigg)!~$ This is based on

the fact that $n!=\mathcal G\bigg(\dfrac1n\bigg)$, where $\mathcal G(n)=\displaystyle\int_0^\infty~e^{-t^n}~dt.~$ See $\Gamma$ function for more details.

I don't know anything about the gamma function and am not allowed to use it.

You can't not use it, since, as you can clearly see, there is no alternate way of expressing the result.

Integration of $\int_0^\infty \sqrt x e^{-x^2} dx$

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