Integrate $\int_{-\infty}^{\infty} \frac{1}{(1-jt)^{N-1} (1-jat)} e^{-j t x} dt$ I would like to solve the following integral:
$$ \int_{-\infty}^{\infty} \frac{1}{(1-jt)^{N-1} (1-jat)} e^{-j t x} dt $$
Where $j = \sqrt{-1}$, $N \in \mathbb{Z}_{++}$, $a \in \mathbb{R}_{++}$ and $a \neq 1$.

So it appears I will need to use residue calculus here. I think there are singularities at:
$$ t = -j \quad \text{and} \quad t = -\frac{1}{a} j$$
However, I'm not sure where to draw my contour, since as $a \rightarrow 0$, the singularity moves towards $-j \infty$. 

Can anyone provide a solution, or an approach?
Thanks

Partial Answer:
We will first consider the case $N=1$. In this case, making the sign switch as Zachary pointed out, we have the integral: 
$$ \int_{-\infty}^{\infty} \frac{1}{(1+jat)} e^{j t x} dt $$
If we make an arc in the upper half plane, we create the contour integral:
$$ \int_{\Gamma} \frac{1}{(1+jaz)} e^{j x z} dz = \int_{-\infty}^{\infty} \frac{1}{(1+jaz)} e^{jxz} dz + \int_{C_{R}} \frac{1}{(1+jaz)} e^{j x z} dz = j 2 \pi  Res \Big\{ z = \frac{j}{a} \Big \} $$
We evaluate the integral on the arc first. Letting $z = r e^{j \theta}$, we get:
$$ | \int_{C_{R}} \frac{1}{(1+jaz)} e^{j x z} dz | \leq  \int_{C_{R}} | \frac{1}{(1+jaz)} e^{j x z} dz| \leq \int_{C_{R}} \frac{1}{|1+jaz|} |e^{j x z}| |dz| \leq \int_{0}^{2 \pi} \frac{1}{|1+jaR e^{j \theta}|} R d \theta \leq \int_{0}^{2 \pi} \frac{1}{{a^{2}R^{2} e^{j 2\theta}}} R^{2} d \theta \leq  \frac{1}{a^{2}}  \int_{0}^{2 \pi} e^{-j 2 \theta} d\theta = 0$$
This leaves us with:
$$ \int_{-\infty}^{\infty} \frac{1}{(1+jaz)} e^{jxz} dz  = j 2 \pi Res \Big\{ z = \frac{j}{a} \Big \} $$
The residue can be computed via:
$$ Res \Big\{ z = \frac{j}{a} \Big \} = \frac{e^{j x z}}{\frac{d}{dz} (1+jaz)} \Big |_{z=\frac{j}{a}} = \frac{e^{-\frac{x}{a}}}{ja} $$
Thus:
$$ \int_{-\infty}^{\infty} \frac{1}{(1+jaz)} e^{jxz} dz  = 2 \pi \frac{1}{a} e^{-\frac{x}{a}} $$
Which agrees with Zachary's answer.

Now for general $N$. 
We have the contour:
$$ \int_{\Gamma} \frac{1}{((1+jz)^{N-1}) (1+jaz)} e^{j x z} dz = \int_{-\infty}^{\infty} \frac{1}{(1+jz)^{N-1} (1+jaz)} e^{jxz} dz + \int_{C_{R}} \frac{1}{(1+jz)^{N-1}(1+jaz)} e^{j x z} dz = j 2 \pi  \big( Res \Big\{ z = \frac{j}{a} \Big \} + Res \Big\{ z = {j} \Big \} \big) $$
We can evaluate the contour on the arc, via:
$$ | \int_{C_{R}} \frac{1}{(1+jz)^{N-1} (1+jaz)} e^{j x z} dz | \leq \int_{0}^{2 \pi} \frac{1}{|(1+jR e^{j \theta})^{N-1} (1+jaR e^{j \theta})|} R d\theta  \leq 2\pi \frac{1}{(1+R^{2})^{N-1} (1+a^{2}R^{2})|} R^{2} \rightarrow 0  $$
Thus: 
$$ \int_{-\infty}^{\infty} \frac{1}{(1+jz)^{N-1} (1+jaz)} e^{jxz} dz  = j 2 \pi \big( Res \Big\{ z = \frac{j}{a} \Big \} + Res \Big\{ z = {j} \Big \} \big) $$
Now we want to calculate $Res \Big\{ z = \frac{j}{a} \Big \}$. We use:
$$ Res \Big\{ z = \frac{j}{a} \Big \} = \underset{z \rightarrow z_{0}}{lim} (z-z_{0}) f(z)$$
We let $f(z) = \frac{1}{(1+jz)^{N-1} (1+jaz)} e^{jxz}$. However, first we need to rewrite it. We notice that:
$$ (1+jaz) = ja(z-\frac{j}{a})$$
Thus:
$$ \underset{z \rightarrow z_{0}}{lim} (z-z_{0}) f(z) = \underset{z \rightarrow z_{0}}{lim} (z-z_{0}) (z-\frac{j}{a})  \frac{1}{ja(1+jz)^{N-1} (z-\frac{j}{a})} e^{jxz} = \frac{e^{-\frac{x}{a}}}{ja (1-\frac{1}{a})^{N-1}}$$
This agrees with the special case result of $N=1$.

Now we want to calculate $Res \Big\{ z = \frac{j}{a} \Big \}$. We use:
$$ Res \Big\{ z = j \Big \} = \frac{1}{(N-2)!} \underset{z \rightarrow z_{0}}{lim} \frac{d^{N-2}}{z^{N-2}}(z-z_{0})^{N-1} f(z)$$
To start, we again rewrite our $f(z)$ noticing that:
$$ (1 + jz)^{N-1} = j^{N-1}(z-j)^{N-1} $$
Therefore, our problem becomes to find the following derivative and evaluate it at $z=j$
$$\frac{1}{(N-2)!} \frac{d^{N-2}}{z^{N-2}} \frac{e^{jxz}}{j^{N-1}(1+jaz)}$$
Which is equivalent to:
$$\frac{j^{1-N}}{(N-2)!} \frac{d^{N-2}}{z^{N-2}} \frac{e^{jxz}}{(1+jaz)}$$
We use the Generalized Leibniz rule, with $f = e^{jxz}$ and $g = (1+jaz)^{-1}$. The derivatives are:
$$ f^{(n-k)} = j^{n-k} x^{n-k} e^{jxz} $$
$$ g^{(k)} = (-1)^{k \downarrow} j^{k} a^{k} (1+jaz)^{-(k+1)} $$
Where $(-1)^{k \downarrow} = \prod_{i=0}^{k} (i-1)$ is the falling factorial function.
According to the general Leibniz rule then, we have:
$$ (fg)^{(n)} = \sum_{k=0}^{n} \binom{n}{k} f^{(n-k)} g^{(k)} $$
This gives us:
$$ (fg)^{(N-2)} = j^{N-2} \sum_{k=0}^{N-2} (-1)^{k \downarrow} \binom{N-2}{k} x^{N-2-k} \frac{a^{k} e^{jxz}}{(1+jaz)^{k+1}} $$
Thus our residue is:
$$ Res \Big\{ z = j \Big \} = \underset{z \rightarrow z_{0}}{lim} \frac{j^{1-N}}{(N-2)!} j^{N-2} \sum_{k=0}^{N-2} (-1)^{k \downarrow} \binom{N-2}{k} x^{N-2-k} \frac{a^{k} e^{jxz}}{(1+jaz)^{k+1}} $$
$$ Res \Big\{ z = j \Big \} = \frac{1}{j(N-2)!} \sum_{k=0}^{N-2} (-1)^{k \downarrow} \binom{N-2}{k} x^{N-2-k} \frac{a^{k} e^{-x}}{(1-a)^{k+1}} $$
Is this correct?
 A: Your integral is equivalently (I am using $i=\sqrt{-1}$)
$$\int_{-\infty}^\infty dt \frac{e^{ixt}}{(1+it)^{N-1}(1+iat)}$$
Your integrand now has poles at $t=i$ and $t=i/a$. If you integrate your function on the classic semi-circular contour in the upper-half plane, you should find that the integral on the semi-circle vanishes as its radius $R\to\infty$, and you are left with the integral above. 
Then, by the Residue Theorem, your integral is just the sum of the two residues of the integrand multiplied by $2\pi i$.
The residue at the pole $t=i/a$ should be
$$\frac{a^{N-2}e^{-x/a}}{i(a-1)^{N-1}}$$
The other one may be a little difficult because it is of order $N-1$. 
The $a\to 0$ limit shouldn't be a problem because since we have left the radius $R$ of the semi-circle go to infinity, your pole at $t=i/a$ would be contained in it for any $a>0$.
Continued Answer:
The residue for the second pole at $t=i$ can be shown to be 
$$\frac{2\pi}{i^{N-2}(N-2)!}\lim_{z\to i}\left(\frac{d}{dz}\right)^{N-2}\frac{e^{ixz}}{1+iaz}$$
Using the generalized Leibniz rule, we can show that
$$\left(\frac{d}{dz}\right)^{N-2}\frac{e^{ixz}}{1+iaz}=(iz)^{N-2}\sum_{k=0}^{N-2}(-1)^k {N-2\choose k}x^{N-2-k} \frac{a^k}{(1+iaz)^{k+1}}$$
Summing the residues and taking their product with $2\pi i$, your integral is
\begin{align}
\frac{2\pi}{(a-1)^{N-1}}a^{N-2}e^{-x/a}+\frac{2\pi}{i^{N-2}(N-2)!}(-1)^{N-1}\sum_{k=0}^{N-2}{N-2\choose k} x^{N-2-k}\frac{a^k}{(a-1)^{k+1}}
\end{align}
