Integration $\int_{0}^{\infty}\frac{\sin(ax)}{x^2}e^{-zx}\text dx$ I need help solving, 
$$\int_{0}^{\infty}\frac{\sin(ax)}{x^2}e^{-zx}\text dx$$
My attempt resulted in
$$\int_{0}^{\infty}\frac{\sin(ax)}{x^2}e^{-zx}\text dx = \frac{a}{2}\arctan\left(\frac{a}{z}\right) + z \ln(z^2 + a^2)$$
I doubt this is correct because when I apply it to my research of computing charge distributions I get wrong results. Thank you.

You can view this integral as the Mellin transform of $\sin(a x)\exp(-z x)$ at a certain value of $s$. Using Ramanujan's master theorem as a method we have the following:
Write the series expansion of the 
$$
\sin(a x)\exp(-z x) = \sum_{k=0}^\infty \phi(k) \frac{(-x)^k}{k!}
$$
where 
$$
\phi(k)=\frac{i(-1)^k}{2}\left((-ia-z)^k-(ia-z)^k\right)
$$
so the Mellin transform is given by
$$
\int_0^\infty x^{s-1}\sin(a x)\exp(-zx)\;dx = \Gamma(s)\phi(-s)
$$
if we take $s=1$ then we have 
$$
\int_0^\infty \frac{\sin(a x)\exp(-zx)}{x^2}\;dx = \phi(-1) = \frac{a}{a^2+z^2}
$$
the potential problem with this is that we assumed $\phi(k)$ could be extended to take negative values.
