Answered: dx L2 exp %|

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Step 1

Given integral is $I = \int_{- \infty}^{\infty} e^{(- \frac{x^{2}}{L^{2}})} d x$

This type of integral cannot be done using elementary integration techniques. So let us convert Cartesian coordinates to polar coordinates form and try to evaluate the integral.

we can notice that, $I = \int_{- \infty}^{\infty} e^{(- \frac{y^{2}}{L^{2}})} d y$ in terms of y.

So, now let us try to evaluate double integral $I^{2}$ with x and y terms.

$\begin{array}{rcl} I^{2} & = & \int_{- \infty}^{\infty} e^{(- \frac{x^{2}}{L^{2}})} d x \cdot \int_{- \infty}^{\infty} e^{(- \frac{y^{2}}{L^{2}})} d y \\ = & \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{(- \frac{y^{2}}{L^{2}})} e^{(- \frac{x^{2}}{L^{2}})} d y d x \\ = & \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{(- \frac{(x^{2} + y^{2})}{L^{2}})} d y d x \end{array}$

Let us now convert x-y coordinates into polar coordinates, by substituting $x^{2} + y^{2} = r^{2}$

Limits of x and y changes to limits of r and $θ$ , which gives:

limits of r from $(0, \infty)$ and $θ$ from $(0, 2 π)$

Now, writing integral in polar coordinates form:

$\begin{array}{rcl} I^{2} & = & \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{(- \frac{(x^{2} + y^{2})}{L^{2}})} d y d x \\ = & \int_{0}^{2 π} \int_{0}^{\infty} e^{(- \frac{r^{2}}{L^{2}})} r d r d θ \end{array}$

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