Advanced techniques needed to solve a difficult integral. I am looking to solve the following integral.
$\int_{0}^{\infty} \frac{1-\cos(ax)}{x^2}e^{bx} dx$. 
I have made an attempt using the differentiation under the integral sign method and I got the following:
$b\ln(b)-\frac{1}{2}b-\frac{1}{2}b\ln(b^2+a^2)-\frac{1}{2}a\arctan(\frac{b}{a})$. eqn(2)
I think the above answer is incorrect because I know that if b=0, then
$\int_{0}^{\infty} \frac{1-\cos(ax)}{x^2} dx = \frac{\pi}{2}|a|$
however, if I let b$\rightarrow$0 in  eqn(2) my result is 
$\int_{0}^{\infty} \frac{1-\cos(ax)}{x^2}e^{bx} dx = 0$.
I would appreciate any assistance provided. Thank you.

According to Mathematica the answer for $b<0$ and $a$ real is :
$$\frac{1}{2}b \log(a^2+b^2) -b \log(-b) - |a|\arctan \left(\frac{|a|}{b}\right)$$
which indeed converges to the value $|a|\frac{\pi}{2}$ when $b \rightarrow 0$. It is closed to you expression but not exactly the same.
What exactly did you do with differentiation under the integral ? For example, using this kind of method, you have to use some inital value to fix your integration constants. Which one did you use ?
