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I encountered the following improper integral: $I=\int\limits_0^\infty {{e^{ - x}}\left| {\sin x} \right|dx}$. I solved the problem as follows: $I=\int\limits_0^\pi {{e^{ - x}}\sin xdx} - \int\limits_\pi ^{2\pi } {{e^{ - x}}\sin xdx} + \int\limits_{2\pi }^{3\pi } {{e^{ - x}}\sin xdx} + ... + {\left( { - 1} \right)^n}\int\limits_{n\pi }^{\left( {n + 1} \right)\pi } {{e^{ - x}}\sin xdx} + ...$ Because: $\int {{e^{ - x}}\sin xdx} = \frac{{ - {e^{ - x}}\left( {\sin x + \cos x} \right)}}{2}$ then we have: $I = \left. {\frac{{ - {e^{ - x}}\left( {\sin x + \cos x} \right)}}{2}} \right|_0^\pi - \left. {\frac{{ - {e^{ - x}}\left( {\sin x + \cos x} \right)}}{2}} \right|_\pi ^{2\pi } + ... + {\left( { - 1} \right)^n}\left. {\frac{{ - {e^{ - x}}\left( {\sin x + \cos x} \right)}}{2}} \right|_{n\pi }^{\left( {n + 1} \right)\pi } + ...$ or: $I = 0.5 + 1\left( {{e^{ - \pi }} + {e^{ - 2\pi }} + {e^{ - 3\pi }} + ...} \right)= 0.5 + \frac{1}{{{e^\pi } - 1}}$ My question is: Is there any better and more general method to solve this problem, that I can apply in some other similar problems?

Viet
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  • I am sorry but it's not an "improper" integral. It would have been if the objective is to compute $I=\int\limits_0^\infty {{e^{ - x}}\sin x dx}$ whereas $I=\int\limits_0^\infty {{e^{ - x}}\left| {\sin x} \right|dx}$ does not converge... – Jean Marie Sep 09 '16 at 08:43
  • In fact, the mentioned integral converges and can be calculated as shown – Viet Sep 09 '16 at 14:54
  • I think that you have misunderstood what I was saying, by my "would have been" ; indeed $I=\int\limits_0^\infty {{e^{ - x}}\left| {\sin x} \right|dx}$ is convergent !! It is why I said that qualifying it as "improper" is ... improper :):):) – Jean Marie Sep 09 '16 at 15:05
  • In the world of Riemann integration this is definitely an improper integral, and it converges to the value found by the OP, which then is declared the value of the integral (similar to the sum of an infinite series). – In the world of Lebesgue integration this integral exists, i.e., its value is a finite number. – Christian Blatter Sep 09 '16 at 15:59
  • Very thorough answer (distinction Riemann/Lebesgue). I realized my error (I had always used the term "generalized" for this kind of integrals). – Jean Marie Sep 09 '16 at 18:33

2 Answers2

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Shifting the argument of the integrand by $\pi$ amounts to multiplying by $-e^{-\pi}$.

Assume you know that

$$J=\int_0^\infty e^{-x}\sin x\, dx=\frac12,$$ which is easily established with $e^{-x}\sin x=\Im{e^{(-1+i)x}}\to J=-\Im(-1+i)^{-1}$.

Then

$$K=J+Je^{-\pi}=\frac12(1+e^{-\pi})$$ is the integral of the first arch (i.e. from $0$ to $\pi$), by cancellation of the other arches. And by summing on all rectified arches (that form a geometric series), the requested integral is

$$I=\frac K{1-e^{-\pi}}=\frac12\frac{1+e^{-\pi}}{1-e^{-\pi}}=\frac12\coth\frac\pi2.$$

enter image description here

  • Nice argumentation, nice drawing. Besides, do you agree with me that it's not an "improper" integral ? Because the OP doesn't agree with this idea ... – Jean Marie Sep 09 '16 at 15:13
  • @JeanMarie The integral is indeed called improper because the upper bound does not exist, and it should be written as a limit. This a converging improper integral. –  Sep 09 '16 at 16:12
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    You are right, I would have at first called such an integral a "generalized integral" but thinking more about it, "improper" is the good adjective. – Jean Marie Sep 09 '16 at 18:31
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Note that by letting $t=x-k\pi$, we have that $$\int_{k\pi}^{(k+1)\pi}e^{-x}|\sin(x)|dx=\int_{0}^{\pi}e^{-(t+k\pi)}|\sin(t+k\pi)|dx\\=e^{-k\pi}\int_{0}^{\pi}e^{-t}|\sin(t)| dt=e^{-k\pi}\int_{0}^{\pi}e^{-t}\sin(t) dt\\=e^{-k\pi}\cdot\left[- \frac{e^{ -t}}{2}(\sin t + \cos t) \right]_0^{\pi}=e^{-k\pi}\cdot\frac{1+e^{-\pi}}{2}.$$ Hence $$I=\frac{1+e^{-\pi}}{2}\sum_{k\geq 0}e^{-k\pi}=\frac{1+e^{-\pi}}{2(1-e^{-\pi})}=\frac{1}{2} + \frac{1}{{{e^\pi } - 1}}$$ which now is the same of your edited result.

Robert Z
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