# Integrating $\int^{\infty}_0 e^{-x^2}\,dx$ using Feynman's parametrization trick

I stumbled upon this short article on last weekend, it introduces an integral trick that exploits differentiation under the integral sign. On its last page, the author, Mr. Anonymous, left several exercises without any hints, one of them is to evaluate the Gaussian integral $$\int^\infty_0 e^{-x^2} \,dx= \frac{\sqrt{\pi}}{2}$$ using this parametrization trick. I had been evaluating it through trial and error using different paramatrizations, but no luck so far.

Here are what I have tried so far:

• A first instinct would be do something like:$$I(b) = \int^\infty_0 e^{-f(b)x^2}\,dx$$ for some permissible function $f(\cdot)$, differentiating it will lead to a simple solvable ode: $$\frac{I'(b)}{I(b)} = -\frac{f'(b)}{2f(b)}$$ which gives: $$I(b) = \frac{C}{\sqrt{f(b)}}.$$ However, finding this constant $C$ basically is equivalent to evaluating the original integral, we are stuck here without leaving this parametrization trick framework.

• A second try involves an exercise on the same page: $$I(b) = \int^\infty_0 e^{-\frac{b^2}{x^2}-x^2}dx.$$ Taking derivative and rescaling the integral using change of variable we have: $$I'(b) = -2I(b).$$ This gives us another impossible to solve constant $C$ in: $$I(b) = C e^{-2b}$$ without leaving this framework yet again.

• The third try is trying modify Américo Tavares's answer in this MSE question: $$I(b) = \int^\infty_0 be^{-b^2x^2}\,dx.$$ It is easy to show that: $$I'(b) = \int^\infty_0 e^{-b^2x^2}\,dx - \int^\infty_0 2b^2 x^2 e^{-b^2x^2}\,dx = 0$$ by an integration by parts identity: $$\int^\infty_0 x^2 e^{- c x^2}\,dx = \frac{1}{2c}\int^\infty_0 e^{- c x^2}\,dx .$$ Then $I(b) = C$, ouch, stuck again at this constant.

Notice in that Proving $\displaystyle\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$ question, Bryan Yocks's answer is somewhat similar to the idea of parametrization, however he has to introduce another parametric integration to produce a definite integral leading to $\arctan$.

Is there such a one shot parametrization trick solution like the author Anonymous claimed to be "creative parameterizations and a dose of differentiation under the integral"?

• I am 90% sure Bryan Yock's solution is the intended one. It appeared in another article on differentiation under the integral sign I read where the Gaussian integral was computed. You give a good analysis of why a simple 1-shot parameterization will not work. – Potato May 13 '13 at 23:33
• I was wondering that as well when reading the article. math.uconn.edu/~kconrad/blurbs/analysis/gaussianintegral.pdf go to the third proof. it's called differentiation under the integral but it's not using the method that other article introduces. so there might be another way still. – pad May 13 '13 at 23:36
• @Potato any link? – Shuhao Cao May 14 '13 at 1:06
• @pad Thanks for the input! Possibly the author is lurking on MSE and read those answers, haha. – Shuhao Cao May 14 '13 at 1:07
• Honest to God I didn't read Bryan Yock's solution before I found mine! I just looked at it, saw two functions and ran away to work with the single function I'd defined P: Hope my answer is a little more satisfying. – Sharkos May 14 '13 at 1:13

Just basically independently reinvented Bryan Yock's solution as a more 'pure' version of Feynman.

Let $$I(b) = \int_0^\infty \frac {e^{-x^2}}{1+(x/b)^2} \mathrm d x = \int_0^\infty \frac{e^{-b^2y^2}}{1+y^2} b\,\mathrm dy$$ so that $I(0)=0$, $I'(0)= \pi/2$ and $I(\infty)$ is the thing we want to evaluate.

Now note that rather than differentiating directly, it's convenient to multiply by some stuff first to save ourselves some trouble. Specifically, note

$$\left(\frac 1 b e^{-b^2}I\right)' = -2b \int_0^\infty e^{-b^2(1+y^2)} \mathrm d y = -2 e^{-b^2} I(\infty)$$

Then usually at this point we would solve the differential equation for all $b$, and use the known information at the origin to infer the information at infinity. Not so easy here because the indefinite integral of $e^{-x^2}$ isn't known. But we don't actually need the solution in between; we only need to relate information at the origin and infinity. Therefore, we can connect these points by simply integrating the equation definitely; applying $\int_0^\infty \mathrm d b$ we obtain

$$-I'(0)= -2 I(\infty)^2 \quad \implies \quad I(\infty) = \frac{\sqrt \pi} 2$$

• I love this answer! Of course, +1. – Sangchul Lee May 14 '13 at 1:45
• +1 for this clear and concise answer. And apology for the unnecessary edit. -Mark – Mark Viola May 31 '16 at 15:23
• Wait why do we know that $$\frac{\partial}{\partial b}\frac{I(b)}{be^{b^2}}=-2e^{-b^2}I(\infty)$$? – clathratus Nov 6 '18 at 16:47
• @clathratus Use the second form of $I(b)$. Firstly simplify $I \exp(-b^2)/b$. Then it should be clear why the derivative is simple to compute. – Sharkos Nov 6 '18 at 16:51