solving $\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{e^{ikx}dk}{b^2+k^2}$

Question

Can someone explain to me how to solve this integral?

I didn't understand the method how to approach it in the class:

$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{e^{ikx}dk}{b^2+k^2}$$

Is there any identity that i should know about?

Thanks

Answer 1

If you were to solve $$ -f''+b^2f = \sqrt{2\pi}\delta $$ for a function $f\in L^2$, where $\delta$ is the Dirac delta, then you would end up with the equation $$ k^2\hat{f}(k)+b^2\hat{f}(k)=1 \\ \hat{f}(k)=\frac{1}{k^2+b^2} \\ f(x)= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(k)e^{ikx}dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{e^{ikx}}{k^2+b^2}dk, $$ which is the desired integral.

The differential equation is solved by $$ f(x)=Ce^{-b|x|}, $$ where the discontinuity in $(-f')$ is chosen to have a jump discontinuity of $\sqrt{2\pi}$, so that the derivataive of $-f'$ will produce $\sqrt{2\pi}\delta$. So, $C$ must be chosen such that $$ -\sqrt{2\pi}=f'(0+)-f'(0-)=(-b)C-(b)C=-2bC \implies C=\frac{\sqrt{2\pi}}{2b} $$ So the expected solution is $$ f(x) = \frac{\sqrt{2\pi}}{2b}e^{-b|x|}. $$ To check this, $$ \hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\sqrt{2\pi}}{2b}e^{-b|x|}e^{ikx}dx \\ =\frac{1}{2b}\left(\int_{-\infty}^{0}e^{bx}e^{ikx}dx+\int_{0}^{\infty}e^{-bx}e^{ikx}dx\right) \\ = \frac{1}{2b}\left(\frac{1}{b+ik}-\frac{1}{-b+ik}\right) \\ = \frac{1}{k^2+b^2}. $$ This implies $$ \frac{\sqrt{2\pi}}{2b}e^{-b|x|}= f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(k)e^{ikx}dx = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{1}{k^2+b^2}e^{ikx}dk, $$ which evaluates your integral.