6
$\begingroup$

Problem: Prove that $$I = \int_{0}^\infty e^{-x^{2}}x^{2n+1}dx = \frac{n!}{2} $$

Source: A problem I found on an integral test. The problem bugged me for long and I did end up leaving it on the exam. Back home I decided to tackle it again. Here's my go on it.

My try: I have never encountered such a problem before, not even in my assignments and workbooks. Here's how I tried it

We can first take an indefinite integral and then we can work up towards a reduction formula.We can plug in the limits later. I don't know if it's any good but I did write this up on the exam: $$I_{2n+1} = \int{e^{x^{-2}}}{x^{2n+1}}dx$$

On applying Integration by parts:

$$I_{2n+1} = e^{-x^{2}}\frac{x^{2n+2}}{2n+2} +\frac{2}{2n+2} \int{e^{-x^{2}}}{x^{2n+3}}dx$$

or $$I_{2n+1} = e^{-x^{2}}\frac{x^{2n+2}}{2n+2} +\frac{I_{2n+3}}{n+1} $$ or

$$I_{2n+1} = {e^{-x^{2}}\over 2}\Bigl(\frac{x^{2n+2}}{n+1}\Bigl) +\frac{I_{2n+3}}{n+1} $$

What can I do next? Am I doing it incorrectly? Is this the wrong way? Please help me as I can't stop thinking about this problem! Thanks!

Edit 1: As suggested (already tried by me) lets put $u = -x^2$ $$ du = -2xdx$$ plug the above expression in the problem

$$I_{n} = -{1 \over 2}\int{e^{u}}{u^{n}}du$$

$$I_{n} = -{1 \over 2}e^{-u}\frac{u^{n+1}}{n+1} + \frac{I_(n+1)}{n+1}$$

Certainly helps. Now what?

Edit 2: Use the gamma function to get the result. Now unfortunately I didn't read it for the test that's why I didn't get it. Thank you all for pointing out the right direction!

$\endgroup$
6
  • 2
    $\begingroup$ $$e^{x^{-2}}$$ or $$e^{-x^2}\ ?$$ $\endgroup$
    – Did
    Sep 19, 2017 at 15:42
  • $\begingroup$ The latter is the arch classical Gaussian integral. Tons of questions about it here. $\endgroup$
    – Did
    Sep 19, 2017 at 15:43
  • $\begingroup$ I'm very sorry I'll correct it now @Did thanks for pointing out $\endgroup$ Sep 19, 2017 at 15:45
  • 1
    $\begingroup$ Try the substitution $t = x^2$ on your initial integral $\endgroup$ Sep 19, 2017 at 15:54
  • 1
    $\begingroup$ First to simplify perform the change of variable $y=x^2$ $\endgroup$
    – FDP
    Sep 19, 2017 at 15:54

4 Answers 4

7
$\begingroup$

I thought it might be instructive to present a way forward that relies on differentiating under the integral. To that end, we proceed.


Let $I(a)$ be given by the integral

$$I(a)=\int_0^\infty e^{-ax^2}\,x\,dx=\frac1{2a} \tag1$$

Differentiating $(1)$ $n$ times reveals

$$\begin{align} I^{(n)}(a)&=(-1)^n\int_0^\infty e^{-ax^2}\,x^{2n+1}\,dx\\\\ &=\frac12 \frac{d^n}{da^n}\left(\frac1a\right)\\\\ &=\frac1{2a^{n+1}}(-1)^n\,n!\tag2 \end{align}$$

Finally, setting $a=1$ in $(2)$, we find that

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty e^{-x^2}\,x^{2n+1}\,dx=\frac12\,n!}$$

And we are done!

$\endgroup$
2
  • $\begingroup$ Should you equation (2) be $\frac{1}{2}(-1)^n \frac{n!}{a^{n+1}}$? $\endgroup$
    – user88319
    Sep 19, 2017 at 21:32
  • $\begingroup$ @strants Indeed. Good catch. I've edited. Thank you. $\endgroup$
    – Mark Viola
    Sep 19, 2017 at 22:21
2
$\begingroup$

$n\geq 0$ integer.

$\displaystyle J=\int_{0}^\infty e^{-x^{2}}x^{2n+1}dx$

Perform the change of variable $y=x^2$,

$\begin{align}J&=\frac{1}{2}\int_{0}^\infty e^{-x}x^{n}dx\\ &=\frac{1}{2}\Gamma(n+1)\\ &=\boxed{\frac{1}{2} n!}\\ \end{align}$

$\endgroup$
2
  • $\begingroup$ That's the gamma formula you used if I'm not mistaken? $\endgroup$ Sep 19, 2017 at 16:16
  • 1
    $\begingroup$ Sure. It's derived from the formula, $\Gamma(z+1)=z\Gamma(z)$ and using recurrence. $\endgroup$
    – FDP
    Sep 19, 2017 at 16:40
2
$\begingroup$

Let's pretend that we have no idea about the Gamma function. Apply a substitution,

$$t = x^2$$

From which, $$I_n = \int_0^\infty e^{-t} (x^2)^n \cdot \frac x {2x} \mathrm d t$$ $$I_n = \frac 1 2 \int_0^\infty e^{-t} t^n \mathrm dt$$

As we are trying to prove that $2I_n = n!$, it is sufficient to prove that, from the recursive definition of the factorial,

$$2I_1 = 1$$ $$I_n = nI_{n-1}$$

Can you take it from here with IBP?

$\endgroup$
2
  • $\begingroup$ thanks but I don't get the last factorial part. Any references? $\endgroup$ Sep 19, 2017 at 16:19
  • $\begingroup$ Apply IBP to $I_{n+1}$ with $f(x)=exp(-t)$, antiderivative $=-exp(-t)$ and $g(t)=t^{n+1}$ $\endgroup$
    – FDP
    Sep 19, 2017 at 17:47
2
$\begingroup$

Let us start from your initial formula $$ I_{2n+1} = \int{e^{x^{-2}}}{x^{2n+1}}dx $$ and define the definite integral $$ J_{2n+1} = \int_0^{\infty}{e^{x^{-2}}}{x^{2n+1}}dx$$

If we take the formula you got via integration by parts:

$$I_{2n+1} = e^{-x^{2}}\frac{x^{2n+2}}{2n+2} +\frac{2}{2n+2} \int{e^{-x^{2}}}{x^{2n+3}}dx$$ and evaluate it at it's limits, it reduces to

$$J_{2n+1} = \frac{1}{n+1} \int_0^{\infty}{e^{-x^{2}}}{x^{2n+3}}dx=\frac{J_{n+3}}{n+1}.$$

Your formula now follows from evaluating $J_1$ and induction. For this purpose, it is useful to rewrite the equation as $J_{2k+1}=kJ_{2(k-1)+1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.