Evaluate the integral $\int_0^\infty \frac{\sin(ax)}{x} \frac{\sin(bx)}{x}e^{-kx}dx$ I would like to evaluate the following integral:

$$\int_0^\infty \frac{\sin(ax)}{x} \frac{\sin(bx)}{x}e^{-kx}dx$$

I have made several attempts using some crude methods, the workout is way too long to format and type out, so I will make it as brief as possible.
Firstly, I evaluated the integral:

$$\int_0^\infty \frac{\sin(ax)}{x} \frac{\sin(bx)}{x}dx$$

Using integration by parts and the properties of definite integrals:

Setting $u=\sin(ax)\sin(bx)$ and $dv=x^{-2}$

Using:

*

*$\int_{-\infty}^\infty \frac{g(ax)}{x}dx = \int_{-\infty}^\infty \frac{g(x)}{x}dx$

*$\int_{-\infty}^\infty f(x)dx=\frac 1 2\int_{0}^\infty f(x)dx$

*$2\sin a\cos b=\sin(a+b)+\sin(a-b)$
I obtained:

$$\frac1 2b\pi$$

I tried the same approach with the original question, and obtained 3 sets of integrals:


*

*$\frac{1}{2}(a+b)\int_{0}^\infty \operatorname{sinc}xe^{-kx}dx = \frac{1}{2}(a+b)\arctan(k^{-1})$


*$\frac{1}{2}(b-a)\int_{0}^\infty \operatorname{sinc} xe^{-kx}dx = \frac{1}{2}(b-a)\arctan(k^{-1})$


*$(-\frac{1}{2}k)\int_{0}^\infty \frac {\sin(ax)\sin(bx)e^{-kx}}{x}dx$
(These 3 are all summed up)

For (1) and (2), the term $a$ cancels out as expected.
I have no idea how (1) and (2) resulted in $\arctan(k^{-1})$, I just put the intrgral in wolfram alpha. I think inverse Fourier Transforms were involved here. I also don't think $\int_{-\infty}^\infty \frac{g(ax)}{x}dx = \int_{-\infty}^\infty \frac{g(x)}{x}dx$ can be used here, since we have an $e^{-kx}$ term. I went with it just to see where I can get to.
I am unable to solve integral (3).
My second attempt involved dominant and monotone convergence theorems, which was a mess. I don't even know what I was trying to do, hence I will not type it out here. I'm 100% sure it is totally wrong.
We were given some hints:

There's a common and smart trigonometric identity involved, then invoke convergence theorems. The final answer looks very neat.

I suspect the "smart trigonometric identity" is the one listed above, and my answer for the integral without the exponent did indeed look very neat. However using the same method, I wasn't able to obtain the result I needed. It should be a lot easier if convergence theorems are used, at least I assume it would.
I appreciate any help or hints. I'd like to know how this was obtained:

$\int_{0}^\infty \operatorname{sinc} xe^{-kx}dx = \arctan(k^{-1})$

Also please pick out any mistakes I've made, as I am certain my method for the question was highly fallacious and/or mathematically unrigorous. I think I'm on the right track though.
Thanks in advance!
 A: A common strategy is to write $x^{-n}=\frac{1}{\Gamma (n)}\int_0^\infty y^{n-1}e^{-xy}dy$. For example, $$\int_0^\infty\frac{\sin x}{x}e^{-kx}dx=\int_0^\infty dy\Bigg[\int_0^\infty e^{-(k+y)x}\sin x dx\Bigg]=\int_0^\infty\frac{dy}{(k+y)^2+1}.$$
A: I've solved the integral, the answer looks odd, and it is very very long. I'll post the results later. It's 5am the I have a class that begins at 10.
Basically, these are all wrong:


*

*$\frac{1}{2}(a+b)\int_{0}^\infty \operatorname{sinc}xe^{-kx}dx = \frac{1}{2}(a+b)\arctan(k^{-1})$


*$\frac{1}{2}(b-a)\int_{0}^\infty \operatorname{sinc} xe^{-kx}dx = \frac{1}{2}(b-a)\arctan(k^{-1})$


*$(-\frac{1}{2}k)\int_{0}^\infty \frac {\sin(ax)\sin(bx)e^{-kx}}{x}dx$
(These 3 are all summed up)

For 1 and 2, $(a+b)$ cannot be removed from the integral. The result has the inverse tangent function.
For 3, I used dominant convergence theorem which allowed me to differentiate within the integral. (I checked all conditions and they were satisfied) The result was a set of functions involving $ln(x)$.
Here's the the result looks like:

$$\int_0^\infty \frac{\sin(ax)}{x} \frac{\sin(bx)}{x}e^{-kx}dx=\frac{1}{2(a+b)}(\frac{\pi}{2}-arctan(\frac{k}{a+b})+\frac{1}{2(b-a)}(arctan(\frac{k}{b-a})-\frac{\pi}{2})+\frac{barctan(\frac{a-b}{k})}{k}+\frac{1}{2}ln((a+b)^2+k^2)-\frac{1}{2}ln((a-b)^2+k^2)$$

Which is just ridiculous. I assume it can be simplified into something nicer, or I just made a massive mistake somewhere in my 13 pages of writing. I'll know the result later today anyway. I'll post a proper solution if mine is wrong.
