The integral of e^(-x^2) from –infinity to infinity a second way

Up a level : Integrals
Previous page : The integral of e^(-x^2) from –infinity to infinity

The Integral

$I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx}$

in a second way. Ok, could we possibly try a similar trick as with some of the other integrals and insert a constant that we then treat as a variable to differentiate? What if we try

$I(p) = \int\limits_{ - \infty }^\infty {{e^{ - p{x^2}}}dx}$

This would give us

$\begin{gathered} I'(p) = \frac{d}{{dx}}\int\limits_{ - \infty }^\infty {{e^{ - p{x^2}}}dx} = \int\limits_{ - \infty }^\infty {\frac{\partial }{{\partial x}}{e^{ - p{x^2}}}dx} \hfill \\ \quad \quad \; = - \int\limits_{ - \infty }^\infty {{e^{ - p{x^2}}}{x^2}dx} \hfill \end{gathered}$

Perhaps not all that useful. If we try to take partial derivatives then we either need to find the primitive function of our original integrand, or we just get higher and higher powers of x. If we try the substitution u=x² we end up dividing by square root of u instead of multiplying by x².

A second try

Ok, maybe we should try to put our parameter somewhere else. To start with we have that

$I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} = 2\int\limits_0^\infty {{e^{ - {x^2}}}dx}$

Next, we put our parameter as our upper integration limit. We get

$I(p) = 2\int\limits_0^p {{e^{ - {x^2}}}dx}$

Next, we use a similar trick as in the previous method. We square the integral.

$J(p) = {(I(p))^2} = 4{\left( {\int\limits_0^p {{e^{ - {x^2}}}dx} } \right)^2}$

Differentiate with respect to p

Next we take the derivative of this with respect to p– as with some of the previous integrals.

$\begin{gathered} \frac{d}{{dp}}J(p) = 4\frac{d}{{dp}}{\left( {\int\limits_0^p {{e^{ - {x^2}}}dx} } \right)^2} \hfill \\ \quad \quad \quad \;\; = 4 \cdot 2\left( {\int\limits_0^p {{e^{ - {x^2}}}dx} } \right){e^{ - {p^2}}} \hfill \\ \quad \quad \quad \;\; = 8\left( {\int\limits_0^p {{e^{ - {x^2}}}dx} } \right){e^{ - {p^2}}} \hfill \\ \quad \quad \quad \;\; = 8\int\limits_0^p {{e^{ - {x^2}}}{e^{ - {p^2}}}dx} \hfill \\ \quad \quad \quad \;\; = 8\int\limits_0^p {{e^{ - ({x^2} + {p^2})}}dx} \hfill \\ \quad \quad \quad \;\; = 8\int\limits_0^p {{e^{ - {p^2}(\frac{{{x^2}}}{{{p^2}}} + 1)}}dx} \hfill \end{gathered}$

Next, we do the substitution

$\left| \begin{gathered} u = \frac{x}{p} \hfill \\ \frac{{du}}{{dx}} = \frac{1}{p} \hfill \\ dx = pdu \hfill \\ u(0) = \frac{0}{p} = 0 \hfill \\ u(p) = \frac{p}{p} = 1 \hfill \\ \end{gathered} \right.$

This gives us

$\begin{gathered} J'(p) = 8\int\limits_0^p {{e^{ - {p^2}(\frac{{{x^2}}}{{{p^2}}} + 1)}}dx} \hfill \\ \quad \quad \;\; = 8\int\limits_0^1 {{e^{ - {p^2}({u^2} + 1)}}pdu} \hfill \\ \end{gathered}$

Undoing the differentiation

Then we need to undo the differentiation with respect to p, i.e. we need to find

$\int {{e^{ - {p^2}({u^2} + 1)}}pdp}$

To do this we do the substitution

$\left| {\begin{array}{*{20}{c}} {\omega = {p^2}} \\ \begin{gathered} d\omega = 2pdp \hfill \\ \frac{{d\omega }}{2} = pdp \hfill \end{gathered}\end{array}} \right.$

$\begin{gathered} = \frac{1}{2}\int {{e^{ - \omega ({u^2} + 1)}}d\omega } \hfill \\ = - \frac{1}{2}\frac{{{e^{ - \omega ({u^2} + 1)}}}}{{{u^2} + 1}} + C \hfill \\ = - \frac{1}{2}\frac{{{e^{ - {p^2}({u^2} + 1)}}}}{{{u^2} + 1}} + C \hfill \\ \end{gathered}$

Going back to J(p) we thus have

$J(p) = - \frac{8}{2}\int {(\frac{{{e^{ - {p^2}({u^2} + 1)}}}}{{{u^2} + 1}} + C)du}$

Finding the constant of integration

To find C we evaluate this for p=0.

$\begin{gathered} J(0) = - 4\int\limits_0^1 {(\frac{{{e^0}}}{{{u^2} + 1}} + C)du} \hfill \\ \quad \quad = - 4\int\limits_0^1 {(\frac{1}{{{u^2} + 1}} + C)du} \hfill \\ \quad \quad = - 4\left[ {{{\tan }^{ - 1}}u + Cu} \right]_0^1 \hfill \\ \quad \quad = - 4(\frac{\pi }{4} + C) = - \pi - 4C \hfill \end{gathered}$

But on the other hand, we have that

$J(0) = 4{\left( {\int\limits_0^0 {{e^{ - {x^2}}}dx} } \right)^2} = 0$

$- \pi - 4C = 0$

$C = - \frac{\pi }{4}$

Putting it all together

Finally, we let p go toward infinity.

$\begin{gathered} J(\infty ) = \mathop {\lim }\limits_{p \to \infty } \left( { - 4\int\limits_0^1 {(\frac{{{e^{ - {p^2}({u^2} + 1)}}}}{{{u^2} + 1}} - \frac{\pi }{4})du} } \right) \hfill \\ \quad \quad \;\, = - 4\int\limits_0^1 {(\frac{0}{{{u^2} + 1}} - \frac{\pi }{4})du} \hfill \\ \quad \quad \;\, = - 4\int\limits_0^1 {( - \frac{\pi }{4})du} = \int\limits_0^1 \pi = \pi \hfill \end{gathered}$

And we will thus get

$I = \int\limits_{ - \infty }^\infty {{e^{ - {x^2}}}dx} = \sqrt \pi$

I don’t remember where I saw this integral solved in a similar way the first time. I have reproduced it from memory and by reasoning, so I not have done it in completely the same way.

Up a level : Integrals
Previous page : The integral of e^(-x^2) from –infinity to infinity

Last modified: Dec 28, 2023 @ 15:43