Residue of pole

Question

I am trying to integrate $$\frac{1}{2\pi} \int^\infty_{-\infty} \frac{6e^{-ipt}}{(p+1)^2 +9} dp$$

I am using a D-contour and I am trying to calculate the residue at the pole $p = -1$.

I am trying to do this by

$$Residue = 2\pi i\frac{6e^{-ipt}}{2(p+1)}$$

But the denominator still equals $0$ when $p=-1$. Please can someone give me a hint on what I am doing wrong.

Answer 1

There is no pole at $p = -1$. The denominator of the integrand is $9$ there.

The only poles occurs when $(p+1)^2 = -9$.