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