Integrating this complicated integral for statistics

Question

I want to show that :

$$ \int_{-\infty}^{\infty} e^\frac{-u^2}{2} du = \sqrt{2\pi} $$

Is there an elementary way using the tools of Calculus II to do this type of integration? I have not studied numerical analysis yet.

Answer 1

A nice proof is the following.

Let $$\vartheta(t)=\left(\int_0^t e^{-x^2}dx\right)^2$$

Then $$\vartheta'(t)=2\int_0^t e^{-(x^2+t^2)}dx$$

Now let $x=tu$, so that we get $$\vartheta'(t)=2t\int_0^1 e^{-t^2(1+u^2)}du$$ But this gives $$\vartheta (t) = - \int_0^1 {\frac{{{e^{ - {t^2}(1 + {u^2})}}}}{{1 + {u^2}}}} du$$

It follows that the function $$F(t)=\left(\int_0^t e^{-x^2}dx\right)^2+\int_0^1 {\frac{{{e^{ - {t^2}(1 + {u^2})}}}}{{1 + {u^2}}}} du$$ is constant.

Letting $t=0$ we get it is constantly $\dfrac{\pi}4$. As $t\to\infty$ the second integral vanishes, and we get that $$\left(\int_0^\infty e^{-x^2}dx\right)^2=\frac{\pi}4$$

Answer 2

You can use this:

Let $I:= \int e^{-x^2}$. Then: $I^2$= $\int e^{-x^2}dx \int e^{-y^2}dy=\int\int e^{-(x^2+y^2)}dxdy$. Then you can work in polar coordinates, using the fact that $x^2+y^2=r^2 ; x=rcos\theta; y=rsin\theta$, and then find the right region of integration.

Using the substitution:

$I^2= \int_{r=0}^ {\infty}\int_{\theta=0}^{2\pi} e^{-r^2}rdrd\theta$ , gives you the result $2\pi$, so that $I^2=2\pi$ , and then $I= \sqrt{2\pi}$ , which is what you wanted.

Answer 3

The usual trick is to let $I=\int_{-\infty}^\infty e^{-u^2/2}\,du.$ Then $$\begin{align}I^2 &= I\int_{-\infty}^\infty e^{-x^2/2}\,dx\\ &= \int_{-\infty}^\infty e^{-x^2/2}\cdot I\,dx\\ &= \int_{-\infty}^\infty e^{-x^2/2}\int_{-\infty}^\infty e^{-y^2/2}\,dy\,dx\\ &= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-x^2/2}e^{-y^2/2}\,dy\,dx\\ &= \int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)/2}\,dy\,dx\end{align}$$

At that point, we switch to polar coordinates, and note that the plane is covered by $0\le r<\infty$ and $0\le \theta<2\pi,$ so that the new integral is $$I^2=\int_0^{2\pi}\int_0^\infty e^{-r^2/2}\cdot r\,dr\,d\theta=\int_0^\infty e^{-r^2/2}\cdot r\,dr\cdot\int_0^{2\pi}\,d\theta=2\pi\int_0^\infty e^{-r^2/2}\cdot r\,dr.$$

Through a substitution, you should be able to see that $\int_0^\infty e^{-r^2/2}\cdot r\,dr=1,$ so that $I^2=2\pi,$ and so $I=\sqrt{2\pi},$ as desired.

Answer 4

There is indeed such “a way”, but it's not quite “elementary”, and it requires quite a bit of stretch of the imagination, to be quite honest. You must have probably studied factorials and combinations by now, am I correct ? Know then, that it can be proven, for natural numbers, using induction and other primitive methods, that $$n!=\int_0^\infty e^{-\sqrt[n]x}dx$$ and that $$\int_0^1(1-\sqrt[n]x)^mdx=\int_0^1(1-\sqrt[m]x)^ndx=\frac1{C_{m+n}^n}=\frac1{C_{m+n}^m}=\frac{m!\cdot n!}{(m+n)!}$$

Now, let us do something really insane and take the apparently non-sensical case $m=n=\frac12$ . What would we get ? Well, we'd have the following results: $$\tfrac12!=\int_0^\infty e^{-x^2}dx$$ and $$\int_0^1\sqrt{1-x^2}dx=\frac{\frac12!\cdot\frac12!}{\left(\frac12+\frac12\right)!}=\left(\tfrac12!\right)^2$$ But we know that the value of the former integral is $\frac\pi4$ , inasmuch as it describes the area of a quarter of a disc or circle with radius $r=1$ . From which we gather that $$\int_0^\infty e^{-x^2}dx=\tfrac12!=\sqrt\frac\pi4=\frac{\sqrt\pi}2$$ Then we use this result to compute the integral you wanted, by making the simple substitution to the new variable $x=\frac{\sqrt u}2$ .

Answer 5

There is another way which could be used but may be you never heard about this "elementary" function. The antiderivative of your integrand is Sqrt[Pi / 2] Erf[u / Sqrt[2]]. Taking the values at the infinite bounds leads to Sqrt[2 Pi].