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Evaluate

$$\int_{1}^{+\infty}e^{-x}\,\frac{\sin(\alpha x)}{x}\,\mathrm{d}x, \qquad \alpha > 0$$

The integral should converge, because $\sin(x)$ is a bounded function and $\dfrac{1}{x e^x}$ goes to $0$ very fast (much faster than any of the type $\dfrac{1}{x^p}$, where $p$ is arbitrarily large).

I couldn't come up with a substitution. Integration by parts also didn't help. The only thing I see is that $\dfrac{\sin(x)}{x}$ is a special kind of limit:

$$\lim_{x\to0} \frac{\sin(x)}{x} = 1$$

But I don't know how to use that knowledge here.

Integreek
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Egor Okhterov
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  • Have a look at http://math.stackexchange.com/q/159836/321264. – StubbornAtom Aug 13 '16 at 11:09
  • @StubbornAtom Thank you! I think my question should be closed as a duplicate. – Egor Okhterov Aug 13 '16 at 11:15
  • Wait a second... there is no parameter $\alpha$ in a $sin(x)$ in that question :) – Egor Okhterov Aug 13 '16 at 11:18
  • No there isn't. I was merely asking you to look at the general method of solving these integrals for a hint. – StubbornAtom Aug 13 '16 at 11:19
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    As StubbornAtom highlighted, the other question can be generalised to $\sin(\alpha x)$. However, another way to solve your integral is to use $$F(\alpha) = \int_{1}^{\infty} e^{-x} \sin(\alpha x)/x dx$$ and find $F'(\alpha)$, then integrate by parts twice. Alternatively, use that $\sin(\alpha x) = \Im(e^{i \alpha x})$. – Matthew Cassell Aug 13 '16 at 11:19
  • @Mattos Could you, please, elaborate a bit more, how can I take derivative of a function that I don't know? – Egor Okhterov Aug 13 '16 at 11:25
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    @Pixar I'm not sure what you mean? $F(\alpha)$ is the name you have given your integral. Assuming you can differentiate under the integral sign (you can) $$F'(\alpha) = \int_{1}^{\infty} \frac{e^{-x}}{x} \cdot x \cos(\alpha x) dx = \int_{1}^{\infty} e^{-x}\cos(\alpha x) dx$$ – Matthew Cassell Aug 13 '16 at 11:26
  • @Mattos That is quite unexpected. Where can I read about this? Does this fact has a name, so that I can google it? :)

    PS: I meant that we don't have an analytic representation of the function, e.g. $F'(\alpha) = sin(x) cos(x) ln(x) / x^2$.

     – Egor Okhterov Aug 13 '16 at 11:28
  • See here. This technique was used extensively by Feynmann. Also, see here for a proof on when we can differentiate under the integral (Note; read the three conditions in the OPs post which are the requisite conditions for differentiating under the integral. The proof I gave just shows that these conditions are all you need). – Matthew Cassell Aug 13 '16 at 11:30
  • @Mattos Thank you for your help! – Egor Okhterov Aug 13 '16 at 11:32
  • @Mattos It was a random example of what I meant by analytic representation of a function :) I thought that we need to know that representation in order to differentiate the function. – Egor Okhterov Aug 13 '16 at 11:33
  • Sure, just comment if you need more help. – Matthew Cassell Aug 13 '16 at 11:39

1 Answers1

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A general recipe:

1) Derive your integral with respect to $\alpha$. This will produce an $x$ inside your integral that will kill the $x$ in the denominator, leaving you with just $\int _1 ^\infty \Bbb e ^{-x} \sin (\alpha x) \ \Bbb d x$.

2) Integrate by parts twice. If you do it just once it will not be enough. The first integration will turn $\sin$ into $\cos$ and produce some extra factors. The second will turn $\cos$ into $-\sin$ and some further extra factors. You will obtain a 1st degree equation where the unknown is precisely $\int _1 ^\infty \Bbb e ^{-x} \sin (\alpha x) \ \Bbb d x$, that will be trivial to find.

3) Remember that you began by deriving with respect to $\alpha$. Therefore, it is necessary now to integrate the result obtained at (2) with respect to it. This will be an indefinite integral, so will produce an integration constant.

4) Find the value of that integration constant by evaluating the integral given by the problem and your own result at some convenient value of $\alpha$. In this example, choose $\alpha = 0$ (i.e. take the limit toward $0$), because the given integral will be $0$ there.

Alex M.
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