Residue theorem, $\int_{-\infty}^{\infty} e^{-ikx}(1-ika^2)^{-m} dk$ with integer $m$ I am trying to solve this integral $\int e^{-ikx}(1-ika^2)^{-m} dk$ using the residue theorem, but I cannot find the residue of the function.
 $$\frac{1}{(1-ika^2)^{-m}}=\sum (ika^2)^n(-1)^n \binom{-m}{n}$$
But there is not any term of the form $(1-ika^2)$ in the serie expansion. Thank you so much if someone can tell me how to proceed.
 A: To evaluate the integral via the residue theorem, use a semicircular contour in the lower-half complex plane that encloses the pole $z=-i/a^2$.  The integral is then $i 2 \pi$ times the residue at this pole, which is
$$i 2 \pi \frac{1}{(i a^2)^m m!}\left[\frac{d^{m-1}}{dz^{m-1}} e^{-i x z}\right]_{z=-i/a^2} = (-1)^{m} \frac{2 \pi}{a^{2 m}} \frac{x^{m-1}}{m!} e^{-x/a^2}$$
A: $\displaystyle{%
{\rm F}_{m}\left(x, a\right)
\equiv
\int_{-\infty}^{\infty}{\rm e}^{-{\rm i}kx}\,\left(1 - {\rm i}ka^{2}\right)^{-m}\,{\rm d}k}\,,
\quad
m = 0, 1, 2,\ldots$
$$
{\rm F}_{1}\left(x,a\right)
=
{{\rm i} \over a^{2}}\int_{-\infty}^{\infty}
{{\rm e}^{-{\rm i}kx} \over k + {\rm i}/a^{2}}\,{\rm d}k
=
{{\rm i} \over a^{2}}\Theta\left(x\right)\left(-2\pi{\rm i}\right)
{\rm e}^{-{\rm i}\left(-{\rm i}/a^{2}\right)x}
=
{2\pi \over a^{2}}\,\Theta\left(x\right)\,{\rm e}^{-x/a^{2}}\,,
\qquad
a \not= 0
$$
\begin{align}
&\\[5mm]
{\rm F}_{m}\left(x,a\right)
&=
\left\lbrace%
\begin{array}{ll}
2\,\pi\,\delta\left(x\right)\,, &\qquad& \mbox{when}\quad m = 0\ \ \mbox{or}\ \ a = 0
\\[3mm]
{2\pi \over a^{2}}\,\Theta\left(x\right)\,{\rm e}^{-x/a^{2}}
&\qquad&
\mbox{when}\quad m = 1\ \ \mbox{and}\ \ a \not= 0
\\[3mm]
\mbox{diverges} &\qquad& \mbox{when}\quad m = 1\ \mbox{and}\ x = 0
\\[3mm]
0 &\qquad& \mbox{when}\quad m = 2,3,4,\ldots
\end{array}\right.
\end{align}
