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I recently came across Feynman Integration here, thanks to one of Lucian's links I followed from a comment thread today.

It has been quite an interesting read, I must say, but sadly I am (hopelessly) stuck on the concluding exercise. The first question is

$$\int_0^\infty e^{\left(-\frac{x^2}{y^2}-y^2\right)}\,{\rm d}x$$

This already is in two variables, so it looked like I wouldn't have to parameterize it like the examples in the pdf, but differentiating this with respect to $y$ only complicates it further. So it looks like it will have to be parameterized after all.

I would really appreciate if someone explained the thought process behind arriving at a suitable parameterization.

Guy
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1 Answers1

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We can ignore the factor $e^{-y^2}$, and concentrate on

$$F(y) = \int_0^\infty e^{-x^2/y^2}\,dx,$$

where we assume $0 < y < \infty$. Differentiating under the integral then yields

$$F'(y) = \int_0^\infty \frac{2x^2}{y^3}e^{-x^2/y^2}\,dx,$$

and we can apply integration by parts to that:

$$\begin{align} \int_0^\infty \frac{2x^2}{y^3}e^{-x^2/y^2}\,dx &= -\frac{1}{y}\int_0^\infty x\cdot\left(-\frac{2x}{y^2}e^{-x^2/y^2}\right)\,dx\\ &= -\frac{1}{y}\left(\left[xe^{-x^2/y^2}\right]_0^\infty - \int_0^\infty e^{-x^2/y^2}\,dx\right)\\ &= \frac{1}{y}F(y). \end{align}$$

With the initial condition $F(1) = \int_0^\infty e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}$, the differential equation $F'(y) = \frac{1}{y} F(y)$ yields

$$F(y) = \frac{\sqrt{\pi}}{2}y.$$

Adding the ignored factor back in,

$$\int_0^\infty e^{\left(-\frac{x^2}{y^2} - y^2\right)}\,dx = \frac{\sqrt{\pi}}{2}ye^{-y^2}.$$

Daniel Fischer
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