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

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.

  • $\begingroup$ Have a look at math.stackexchange.com/q/159836/321264. $\endgroup$ Aug 13, 2016 at 11:09
  • $\begingroup$ @StubbornAtom Thank you! I think my question should be closed as a duplicate. $\endgroup$ Aug 13, 2016 at 11:15
  • $\begingroup$ Wait a second... there is no parameter $\alpha$ in a $sin(x)$ in that question :) $\endgroup$ Aug 13, 2016 at 11:18
  • $\begingroup$ No there isn't. I was merely asking you to look at the general method of solving these integrals for a hint. $\endgroup$ Aug 13, 2016 at 11:19
  • 1
    $\begingroup$ 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 $\sin(\alpha x) = \Im(e^{i \alpha x})$. $\endgroup$
    – mattos
    Aug 13, 2016 at 11:19

1 Answer 1


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.

  • $\begingroup$ Before taking the derivative, we have to show that the function is continuous and it's derivative is uniformly continuous. How can we prove these things here? For example, $f(x) = \frac{1}{x}$ is not uniformly continuous, so it wouldn't work with that function. $\endgroup$ Aug 16, 2016 at 19:39
  • $\begingroup$ @Pixar: The most used result in these situations is Lebesgue's dominated convergence theorem. $\endgroup$
    – Alex M.
    Aug 16, 2016 at 20:51

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