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 $dv=x^{-2}$


  • $\int_{-\infty}^\infty \frac{g(ax)}{x}dx = \int_{-\infty}^\infty \frac{g(x)}{x}dx$
  • $\int_{-\infty}^\infty f(x)dx=\frac 1 2\int_{0}^\infty f(x)dx$
  • $2\sin a\cos b=\sin(a+b)+\sin(a-b)$

I obtained:

$$\frac1 2b\pi$$

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

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

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

  3. $(-\frac{1}{2}k)\int_{0}^\infty \frac {\sin(ax)\sin(bx)e^{-kx}}{x}dx$ (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!

  • $\begingroup$ Do you know Laplace transform ? $\endgroup$ – C. Dubussy May 14 '18 at 10:51
  • $\begingroup$ Yes, I do know Laplace transforms. $\endgroup$ – Hypergeometry May 14 '18 at 10:55
  • $\begingroup$ For the last one, the antiderivative involves the exponential integral function. $\endgroup$ – Claude Leibovici May 14 '18 at 10:56
  • $\begingroup$ I've never studied exponential integral functions, and I think according to the hint, it isn't required. The answer should look as neat as $\frac{1}{2}b\pi$. I'll look into that function and see what I can spin out of it though. $\endgroup$ – Hypergeometry May 14 '18 at 11:00

A common strategy is to write $x^{-n}=\frac{1}{\Gamma (n)}\int_0^\infty y^{n-1}e^{-xy}dy$. For example, $$\int_0^\infty\frac{\sin x}{x}e^{-kx}dx=\int_0^\infty dy\Bigg[\int_0^\infty e^{-(k+y)x}\sin x dx\Bigg]=\int_0^\infty\frac{dy}{(k+y)^2+1}.$$

  • $\begingroup$ Wow thanks, I didn't think about invoking the gamma function. I'll give it a go. Edit: That worked out nicely. $\endgroup$ – Hypergeometry May 14 '18 at 13:10
  • $\begingroup$ I computed the integral and obtained $\frac{\pi b}{2}$. Still don't know how to solve the third integral in the top post question. $\endgroup$ – Hypergeometry May 14 '18 at 14:28
  • $\begingroup$ @Hypergeometry The integral representation of $1/x$ should work. $\endgroup$ – J.G. May 14 '18 at 15:31
  • $\begingroup$ I used dominant convergence theorem and solved that part. Easy but annoyingly long. $\endgroup$ – Hypergeometry May 15 '18 at 4:41

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:

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

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

  3. $(-\frac{1}{2}k)\int_{0}^\infty \frac {\sin(ax)\sin(bx)e^{-kx}}{x}dx$ (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.

  • $\begingroup$ Update: we were given a hint, and the answer I obtained is likely correct. The due date has been delayed. $\endgroup$ – Hypergeometry May 15 '18 at 4:17

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