- #1

murshid_islam

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- TL;DR Summary
- Attempting to evaluate the Gaussian integral by differentiating under the integral sign

Hi, I have recently learned the technique of integration using differentiation under the integral sign, which Feynman mentioned in his “Surely You’re Joking, Mr. Feynman”. So, I decided to try it on the Gaussian Integral (I do know the standard method of computing it by squaring it and changing to polar coordinates, but I wanted to do it with this method of differentiating under the integral sign). I didn’t manage to do it yet. Here’s what I have tried so far:

I tried to get an ##x## with the ##e^{-x^2}## after differentiating under the integral sign so that I could use the u-substitution ##u = x^2##. But I failed. Then I tried this:

Let ##I(t) = \int_0^{\infty}e^{-tx^2}dx##, where ##t>0## and ##2\cdot I(1) = \int_{-\infty}^{\infty}e^{-x^2}dx##

$$I'(t) = \frac{d}{dt} \int_0^{\infty}e^{-tx^2}dx = \int_0^{\infty} \frac{\partial}{\partial t} \left (e^{-tx^2}\right ) dx = \int_0^{\infty} -x^2 e^{-tx^2}dx$$

Now, I did integration by parts with ##u = -x, du = -dx, dv = xe^{-tx^2}dx, v = \frac{-1}{2t}e^{-tx^2}## and got

$$I'(t) = \frac{-1}{2t}\int_0^{\infty}e^{-tx^2}dx = \frac{-1}{2t} I(t)$$

$$\implies \int \frac{dI}{I} = -\frac{1}{2} \int \frac{dt}{t}$$

$$\implies \ln |I| = \ln \left\lvert \frac{1}{\sqrt{t}}\right\rvert + \ln C = \ln \left\lvert \frac{C}{\sqrt{t}}\right\rvert$$

$$\implies I(t) = \frac{C}{\sqrt{t}}$$

Now I can't find an initial value to use in order to find C. Have I gone down the wrong path?

Let ##I(t) = \int_0^{\infty}e^{-x^2-tx^2}dx## , where ##t>0## and ##2\cdot I(0) = \int_{-\infty}^{\infty}e^{-x^2}dx##

$$I'(t) = \frac{d}{dt} \int_0^{\infty}e^{-x^2-tx^2}dx = \int_0^{\infty} \frac{\partial}{\partial t} \left (e^{-x^2-tx^2} \right ) dx= \int_0^{\infty} (-2tx)e^{-x^2-tx^2} dx = -2t\int_0^{\infty}xe^{-x^2}\cdot e^{-t^2x} dx$$

Now using integration by parts with ##u = e^{-t^2x}, du = -t^2e^{-t^2x}dx, dv = xe^{-x^2}dx, v = \frac{-1}{2}e^{-x^2}## led to this differential equation: ##I'(t) = -t + t^3I(t)##

Any ideas anyone? Am I choosing I(t) wrong?

__First attempt:__I tried to get an ##x## with the ##e^{-x^2}## after differentiating under the integral sign so that I could use the u-substitution ##u = x^2##. But I failed. Then I tried this:

Let ##I(t) = \int_0^{\infty}e^{-tx^2}dx##, where ##t>0## and ##2\cdot I(1) = \int_{-\infty}^{\infty}e^{-x^2}dx##

$$I'(t) = \frac{d}{dt} \int_0^{\infty}e^{-tx^2}dx = \int_0^{\infty} \frac{\partial}{\partial t} \left (e^{-tx^2}\right ) dx = \int_0^{\infty} -x^2 e^{-tx^2}dx$$

Now, I did integration by parts with ##u = -x, du = -dx, dv = xe^{-tx^2}dx, v = \frac{-1}{2t}e^{-tx^2}## and got

$$I'(t) = \frac{-1}{2t}\int_0^{\infty}e^{-tx^2}dx = \frac{-1}{2t} I(t)$$

$$\implies \int \frac{dI}{I} = -\frac{1}{2} \int \frac{dt}{t}$$

$$\implies \ln |I| = \ln \left\lvert \frac{1}{\sqrt{t}}\right\rvert + \ln C = \ln \left\lvert \frac{C}{\sqrt{t}}\right\rvert$$

$$\implies I(t) = \frac{C}{\sqrt{t}}$$

Now I can't find an initial value to use in order to find C. Have I gone down the wrong path?

__Second attempt:__Let ##I(t) = \int_0^{\infty}e^{-x^2-tx^2}dx## , where ##t>0## and ##2\cdot I(0) = \int_{-\infty}^{\infty}e^{-x^2}dx##

$$I'(t) = \frac{d}{dt} \int_0^{\infty}e^{-x^2-tx^2}dx = \int_0^{\infty} \frac{\partial}{\partial t} \left (e^{-x^2-tx^2} \right ) dx= \int_0^{\infty} (-2tx)e^{-x^2-tx^2} dx = -2t\int_0^{\infty}xe^{-x^2}\cdot e^{-t^2x} dx$$

Now using integration by parts with ##u = e^{-t^2x}, du = -t^2e^{-t^2x}dx, dv = xe^{-x^2}dx, v = \frac{-1}{2}e^{-x^2}## led to this differential equation: ##I'(t) = -t + t^3I(t)##

Any ideas anyone? Am I choosing I(t) wrong?

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