# Find a function whose Fourier transform is the following

Find a function whose Fourier transform is the following:

$$\frac{1}{(4+k^{2})(9+k^{2})}$$

I know that $f(x) = F^{-1}\{\hat{f}(k)\}$ so I get:

$$f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(k) e^{ikx}dk$$

I'm unsure about how to solve this integral.. is there a trick that I'm missing?

• Do you know complex analysis? You can do residue theorem. – user223391 Mar 22 '15 at 17:46
• You can use the Fourier transform table. – xpaul Mar 22 '15 at 17:51
• after using partial fractions – Paul Mar 22 '15 at 19:22
• Or the convolution theorem. – Ron Gordon Mar 23 '15 at 23:45

Use complex analysis. Assuming $x$ is positive, we'll close the contour upstairs so that the exponential develops a negative real part and the contribution from the semicircle vanishes. We have a pole at $2i$ and another at $3i$ in the upper half plane, and both are simple so the answer is quick and easy. The answer I am getting is $$\sqrt{2\pi}\cdot\frac{1}{10}\left(\frac{e^{-2x}}{2}-\frac{e^{-3x}}{3}\right).$$ If $x$ is negative, we close downstairs (remembering to account for a sign because the integral is now taken in a clockwise sense). This gives the same, but with $x$ replaced with $-x$, so the inverse Fourier transform is given by the above with $x$ replaced with the absolute value of $x$.
Use partial fractions $$\frac{1}{(4+k^{2})(9+k^{2})}=\frac{1}{5}\left[\frac{1}{4+k^{2}}-\frac{1}{9+k^{2}}\right]$$ I assume these are known to you.