Gaussian integral using Feynman’s technique

6 Jan 2020 Leave a Comment

In my last post we evaluated the following definite integral

$\int_0^\infty e^{-x^t}\mathrm{d}x$

This is the formula we got:

$\int_0^\infty e^{-x^t}\mathrm{d}x=\left (\frac{1}{t} \right )!=\Gamma\left ( \frac{1}{t} +1\right )$

and this is the integral we want to evaluate:

$\int_{-\infty}^{+\infty} e^{-x^2}\mathrm{d}x$

which is equivalent to

$2\int_{0}^{\infty} e^{-x^2}\mathrm{d}x$

because of symmetry: this is an even function, and since the integral has bounds –a and a, it becomes what I’ve just shown.

Here is the graph:

In red: f(x)=e^-x² and area; in blue: F(x)=erf(x)

Since we have the formula, we can simply plug in the 2 and, voilà, we have the answer! We are going to do all of the calculations in a moment, but let’s actually substitute the 2 to see what the result is;

$2\int_0^\infty e^{-x^2}\mathrm{d}x=2\left (\frac{1}{2} \right )!=2\frac{\sqrt\pi}{2}=\sqrt\pi$

If you type 0.5! on Google you’ll get 0.88622692545 , which is the square root of pi. That’s a very beautiful result in my opinion!

But now, let’s actually find the result using Feynman’s technique, or differentiation under the integral sign. I talk about it in this post, where I show how to evaluate the integral from 0 to infinity of sin(x)/x. I am going to use the same approach I used in my previous post, for the generalised form of this integral.

If we had this integral

$2\int_0^\infty xe^{-\alpha x^2}\mathrm{d}x$

an u-substitution would be enough; in order to get that extra x we need to write a new function with a parameter so that, when differentiating with respect to that parameter, the x we need will be generated.

We are going to write the integral as follows:

$I(\alpha)=2\int_0^\infty e^{-\alpha x^2}\mathrm{d}x$

this means that

$2\int_0^\infty e^{-x^2}\mathrm{d}x=I(1)$

we will need this later on, but for now, let’s rewrite the parameterized integral:

$I(\alpha)=2\int_0^\infty e^{-\alpha x^2}\mathrm{d}x$

As we saw here, let’s take the derivative with respect to a on both sides:

$\frac{\mathrm{d} }{\mathrm{d} \alpha}I(\alpha)=2\frac{\mathrm{d} }{\mathrm{d} \alpha}\int_0^\infty e^{-\alpha x^2}\mathrm{d}x$

The derivative can be brought inside the integral, but it becomes partial derivative, since it’s the derivative of each term of the sum; it works the same way as a regular derivative though:

$\frac{\mathrm{d} }{\mathrm{d} \alpha}I(\alpha)=2\int_0^\infty\frac{\partial }{\partial \alpha} e^{-\alpha x^2}\mathrm{d}x$ $I'(\alpha)=2\int_0^\infty -x^2e^{-\alpha x^2}\mathrm{d}x$

Now we can perform a u-substitution:

$u=\alpha x^2\,\,\,\,\,\mathrm{d}u=2\alpha x\,\mathrm{d}x \rightarrow \mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}u$ $I'(\alpha)=-2\int_0^\infty x^2e^{-u}\frac{1}{2\alpha x}\mathrm{d}u$

Don’t forget about the bounds of integration: u=ax², therefore the lower bound is u₁=ax₁²=0 and the upper bound is u₂=ax₂²=infinity. In this case they don’t change, but always remember to do this.

$I'(\alpha)=-\frac{1}{\alpha}\int_0^\infty xe^{-u}\mathrm{d}u$

Since we are integrating with respect to u, we need to write the x in terms of u. We know that u=ax²2, therefore

$u=\alpha x^2\rightarrow x^2=\frac{u}{\alpha}\rightarrow x=\left ( \frac{u}{\alpha} \right )^\frac{1}{2}=u^\frac{1}{2}\alpha^{-\frac{1}{2}}$ $I'(\alpha)=-\alpha^{-1}\int_0^\infty u^\frac{1}{2}\alpha^{-\frac{1}{2}}e^{-u}\mathrm{d}u$ $I'(\alpha)=-\alpha^{-1}\alpha^{-\frac{1}{2}}\int_0^\infty u^\frac{1}{2}e^{-u}\mathrm{d}u$

This is the gamma function! Hence,

$I'(\alpha)=-\alpha^{-1-\frac{1}{2}}\Gamma\left ( \frac{1}{2} +1\right )$

$I'(\alpha)=-\alpha^{-1-\frac{1}{2}}\left ( \frac{1}{2}\right )!$

Now we can take the integral on both sides, since this is I’(a);

$\int I'(\alpha)\,\mathrm{d}\alpha=\int-\left ( \frac{1}{2}\right )!\,\alpha^{-1-\frac{1}{2}}\,\mathrm{d}\alpha$

Let’s apply linearity:

$I(\alpha)=-\left ( \frac{1}{2}\right )!\int\,\alpha^{-1-\frac{1}{2}}\,\mathrm{d}\alpha$

Let’s now apply the power rule:

$I(\alpha)=-\left ( \frac{1}{2}\right )!\frac{\alpha^{-1-\frac{1}{2}+1}}{-1-\frac{1}{2}+1}$ $I(\alpha)=-\left ( \frac{1}{2}\right )!\frac{\alpha^{-\frac{1}{2}}}{-\frac{1}{2}}$ $I(\alpha)=-\left ( \frac{1}{2}\right )!\alpha^{-\frac{1}{2}}\left ( -2 \right )$ $I(\alpha)=2\left ( \frac{1}{2}\right )!\alpha^{-\frac{1}{2}}+C$

Don’t forget the +C!

Overall, we can write

$2\int_0^\infty e^{-\alpha x^2}\mathrm{d}x=2\left ( \frac{1}{2}\right )!\alpha^{-\frac{1}{2}}+C$

Now, we need to find a number that makes the integral equal to 0 in order to find C; after that we will plug in the 1 that we had at the beginning in the original integral and we will find the answer. When a=infinity, we get e^-infinity, which is equal to 0, and therefore the whole integral:

$2\int_0^\infty e^{-\infty}\mathrm{d}x=2\left ( \frac{1}{2}\right )!\cdot \infty^{-\frac{1}{2}}+C$ $0=2\left ( \frac{1}{2}\right )!\cdot 0+C$

When infinity is raised to a negative number, it gives 0; this is why:

$\infty^{-\frac{1}{n}}=\frac{1}{\infty^n}=\frac{1}{\infty}=0$

Anyway,

$0=0+C\rightarrow C=0$

Now, if you remember, we know that

$I(1)=2\int_0^\infty e^{-x^2}\mathrm{d}x$

when a=1; therefore, to find its value, we will plug 1 into this:

$2\int_0^\infty e^{-\alpha x^2}\mathrm{d}x=2\alpha^{-\frac{1}{2}}\left ( \frac{1}{2} \right )!+0$

Hence,

$2\int_0^\infty e^{-x^2}\mathrm{d}x=2\left ( \frac{1}{2} \right )!$

We now know that 0.5!=sqrt(pi)/2, so

$2\int_0^\infty e^{-x^2}\mathrm{d}x=2\frac{\sqrt\pi}{2}$

Finally,

$\int_{-\infty}^\infty e^{-x^2}\mathrm{d}x=\sqrt\pi$

And we’re done!

Hope you liked this post, and if you did give it a like and subscribe to stay updated, it would mean so much to me! If you have any doubts, questions, suggestions, leave a comment and I’ll be happy to reply!

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Integration, Integration techniques, Solutions Calculus, Definite integrals, differentiation under the integral sign, gaussian integral

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Gaussian integral using Feynman’s technique

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