Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to draw the graph of $f(a)=\int_{-\infty}^\infty e^{-ax^2}dx$ on $(0, \infty)$.

I know the graph of $g(x)=e^{-ax^2}$, which is

enter image description here

but I don't know how to graph the integral.

Thank you for your help.

share|improve this question
Hint: change variables to $u = a^{1/2}x$ in the integral –  Zarrax Sep 28 '13 at 1:23

1 Answer 1

up vote 2 down vote accepted

Call the integral $I(a)$. Note that $I(a)$ is undefined for $a\le 0$.

Let $\sqrt{a}\, x=u$. After the substitution, we get $$I(a)=\frac{1}{\sqrt{a}}\int_{-\infty}^\infty e^{-u^2}\,du.$$

So $I(a)$ is a constant times $\dfrac{1}{\sqrt{a}}$. The graph of $y=\dfrac{1}{\sqrt{a}}$ as a function of $a$ is probably familiar.

Detail: From $\sqrt{a}\,x=u$ we get $\sqrt{a}\,dx=du$ and therefore $dx=\frac{1}{\sqrt{a}}du$. Substitute. Then $-ax^2$ becomes $-u^2$, and as $x$ travels from $-\infty$ to $\infty$, so does $u$.

Remark: It turns out that the constant is $\sqrt{\pi}$.

share|improve this answer
Where did $\frac{1}{\sqrt a}$ come from? –  user92324 Sep 28 '13 at 1:32
It comes from the fact that $\int_{-\infty}^\infty e^{-u^2}\,du=\sqrt{\pi}$. There are many proof of this fact, it has been dealt with on MSE repeatedly. The simplest involves calling the integral $J$, and observing that $J^2=\int_{-\infty}^\infty e^{-x^2}e^{-y^2}\,dx\,dy$. Go to polar coordinates, the integral becomes $\int_0^{2\pi} re^{-r^2}\,dr\,d\theta$. This is usually covered in a several variable calculus course. A closely related integral comes up in probability theory, normal distribution. For the graph, you probably don't need the exact value. –  André Nicolas Sep 28 '13 at 1:40
Sorry, I thought you were asking about $\sqrt{\pi}$! When we make the substitution $\sqrt{a}\, x=u$, we get $\sqrt{a}\,dx=du$ and therefore $dx=\frac{1}{\sqrt{a}}\,dx$. I can add this. –  André Nicolas Sep 28 '13 at 1:43
Is it correct to take out $\frac{1}{\sqrt a}$ from the integral even if $a=a(u)$ is a function of $u$?. –  user92324 Sep 28 '13 at 1:50
We are evaluating the integral, treating $a$ as a constant. The result depends on $a$, so is a function of $a$. The integral $\int \frac{1}{\sqrt{a}}e^{-u^2}\,du$ involves only integration with respect to the variable $u$. –  André Nicolas Sep 28 '13 at 1:54

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.