Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have come across integrals of form: \begin{align} &\int\limits_{-\infty}^{+\infty} x\cdot e^{-ax^2} dx\\ &\int\limits_{-\infty}^{+\infty} x^2\cdot e^{-ax^2} dx\\ &\int\limits_{-\infty}^{+\infty} x^3\cdot e^{-ax^2} dx\\ &\int\limits_{-\infty}^{+\infty} x^4\cdot e^{-ax^2} dx\\ \end{align}

Where I have figured out after ploting them that for the ones that have the even exponent ($x^2$, $x^4\dots$) I can write the integral like this:

\begin{align} &2\int\limits_{0}^{+\infty} x^2\cdot e^{-ax^2} dx\\ &2\int\limits_{0}^{+\infty} x^4\cdot e^{-ax^2} dx\\ \end{align}

I have found these integrals in the Bronštein-Semendijajev mathematics manual [page 474] where he states that we can solve them using the formula:

\begin{align} \int\limits_{0}^{\infty}x^n \cdot e^{-ax^2}dx = \frac{1\cdot3\dots(2k-1)\,\,\sqrt{\pi}}{2^{k+1}a^{k+1/2}}\longleftarrow\substack{\text{$n$ is the exponent over $x$}\\\text{while $k=n/2$}} \end{align}

Ok so I can solve these with no problem. But there remains the ones with odd exponent ($x$, $x^3\dots$). On the same page there is a formula for odd exponents, which has a solution:

\begin{align} \int\limits_{0}^{\infty}x^n \cdot e^{-ax^2}dx = \frac{k\text{!}}{2a^{k+1}}\longleftarrow\substack{\text{$n$ is the exponent over $x$}\\\text{while $k=n/2$}} \end{align}

but in my case I have odd functions and I cannot use the relation:

$$\int\limits_{-\infty}^{\infty}dx = 2\int\limits_{0}^{\infty}dx$$

This is why I can't get the form which the mathematical manual needs. When I plotted these even functions I got plots like for example:

enter image description hereenter image description here

From the images I can clearly see that definite integrals between limits $-\infty$ and $\infty$ will equal $0$ for the odd functions.


Graphical solution for the integrals odd functions looks easy while I can't seem to use my mathematics manual to solve them analytically. I am wondering if there is analytical way to show that they equal zero. I was thinking about using relation:

$$\int\limits_{-\infty}^{\infty}dx=\int\limits_{-\infty}^{0}dx + \int\limits_{0}^{\infty}dx$$

somehow. This way I would get similar form that the manual needs, but with swapped integration limits and sign... How do I solve theese?

share|cite|improve this question
+1 for writing effort. But I don't understand what do you mean by analytically solving. The integral of any odd function between symmetric bounds is zero - that's all one needs to say. – Start wearing purple Aug 12 '13 at 11:53
I would like to analytically show, that they equal 0. Check my Edited Question. I was thinking to solve them using the relation described there, but i get spapped limits and a negaitve sign... Long story short, I need to know what are the relations between $$\int\limits_{0}^{\infty} dx \qquad \int\limits_{0}^{-\infty} dx \qquad \int\limits_{\infty}^{0} dx \qquad \int\limits_{-\infty}^{0}$$ – 71GA Aug 12 '13 at 12:01
Hint: once you have decomposed $\int_{-\infty}^{\infty}$ as $\int_{-\infty}^0+\int_0^{\infty}$, look at what happens to the integral $\int_{-\infty}^0$ after the change of variables $x\rightarrow -x$. – Start wearing purple Aug 12 '13 at 12:01
I like hints like this one :) – 71GA Aug 12 '13 at 12:04
I know that $\int_{-\infty}^0 = - \int_{0}^{-\infty}$ but if i want to know what happens if i insert $-x$ instead of $x$ i have to check what function i have. In my case it is odd so i should get the change in sign also... Does this mean that $\int_{-\infty}^{0}=-\int_{\infty}^{0}$ AND $\int_{0}^{\infty}=-\int_{0}^{-\infty}$ ??? – 71GA Aug 12 '13 at 14:17
up vote 2 down vote accepted

For example

$$\int\limits_{-\infty}^\infty xe^{-ax^2}dx=-\frac1{2a}\int\limits_{-\infty}^\infty(2ax\,dx)e^{-ax^2}=\left.-\frac1{2a}e^{-ax^2}\right|_{-\infty}^\infty=0$$

All the rest follow from integrating by parts and/or a little inductive argument.

share|cite|improve this answer
Your case is for the $x$ what about for the $x^2$? – 71GA Aug 12 '13 at 12:03
Told you: integration by parts $$\int x^2e^{-ax^2}dx=-\frac x{2a}e^{-ax^2}+\frac1a\int xe^{-ax^2}=\ldots$$ – DonAntonio Aug 12 '13 at 12:11
One more thing. Isn't it $\int e^{ax^2}dx = \frac{1}{2ax}e^{ax^2}$? – 71GA Aug 12 '13 at 12:17
No. If you multiply by $\,x\,$ the integrand then yes, yet all your exponentials have minus sign in the exponent... – DonAntonio Aug 12 '13 at 12:25
Because the derivative of $\,\frac1{2ax}e^{ax^2}\;$ is not $\;e^{ax^2}\;$ ...! – DonAntonio Aug 12 '13 at 12:54

Let $n \in \mathbb{N}$ an odd number and consider the integral $\int_{\mathbb{R}} x^{n} e^{-ax^{2}} \: dx$. Using the change of variables $t=-x$ ($dt=-dx$), you get :

$$ \int_{\mathbb{R}} x^{n} e^{-ax^{2}} \: dx = - \int_{\mathbb{R}} t^{n} e^{-at^{2}} \: dt$$

So, $\int_{\mathbb{R}} x^{n} e^{-ax^{2}} \: dx = 0$. (I hope I got your question right!)

share|cite|improve this answer
Oh i forgot to mention that $n,~a \in \mathbb{N}$. – 71GA Aug 12 '13 at 12:02

For odd $n$, letting $y = x^2$ so $dy = 2x\ dx$, $\int x^{2m+1}e^{-x^2} dx =(1/2)\int y^m e^{-y} dy $ and this can be done explicitly by repeated integration by parts to get $\int y^m e^{-y} dy =m!\left(1-e^{-y}\sum_{k=0}^m \dfrac{y^k}{k!}\right) $.

For more general $n$, look up "incomplete Gamma function".

share|cite|improve this answer

Use $x = t^{1/2}$. Then

$$ \int_{0}^{\infty}x^{n}{\rm e}^{-x^{2}}\,{\rm d}x = \int_{0}^{\infty}t^{n/2}{\rm e}^{-t}\,{1 \over 2}\,t^{-1/2}\,{\rm d}t = {1 \over 2}\int_{0}^{\infty}t^{\left(n - 1\right)/2}{\rm e}^{-t}\,{\rm d}t = {1 \over 2}\,\Gamma\left(n + 1 \over 2\right) $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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