# How to draw the graph of this function?

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

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

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

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 $\sqrt{\pi}$.

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Where did $\frac{1}{\sqrt a}$ come from? – Twink 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 $u$?. – Twink 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