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.

b bust be negative to ensure the integral convergence. Then if we write the value of the integral we get b in the denominator all through. What you have done is you eliminated the denominators and you know that you want the denominator to be 0. There is no mistake.  
