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I am stuck at this integral. Can anybody help or give me a hint what to do? I have tried to use integration by parts but keep getting stuck.

$$\int_0^\infty \frac{\sin(2x)}{e^x}dx$$

Gary
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Norman
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3 Answers3

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$$\frac{\sin(2x)}{e^x}$$ is the imaginary part of

$$e^{(-1+2i)x},$$

that integrates as

$$\left.\frac{e^{(-1+2i)x}}{-1+2i}\right|_0^\infty=-\frac1{-1+2i}=\frac{1+2i}5.$$

  • I do not know much about integrating complex functions, but this answer looks nice. Why is the downvote? – VIVID Mar 26 '21 at 14:41
  • @VIVID: integrating $e^{(a+bi)x}$ is formally equivalent to simultaneous integration of $e^{ax}\cos(bx)$ and $e^{ax}\sin(bx)$ in terms of themselves, and is just a way of condensing the computation. There is no deep magic, and for the same price you get the cosine integral. Presumably by development was too terse. Thanks for your support :) –  Mar 26 '21 at 14:52
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Write it as

$$\int_0^{+\infty} e^{-x}\sin(2x)\ \text{d}x$$

Integrate by parts getting

$$-\dfrac{1}{2}e^{-x}\cos(2x)\bigg|_0^{+\infty} - \dfrac{1}{2}\int_0^{+\infty} e^{-x}\cos(2)\ \text{d}x$$

The first numerical term goest to $-\dfrac{1}{2}$

Integrate by parts again the second integral to get

$$\dfrac{1}{2}\left(\dfrac{1}{2}e^{-x}\sin(2x) - \int_0^{+\infty} \dfrac{1}{2}e^{-x}\sin(2x)\ \text{d}x\right)$$

The numerical term goest to zero and you are left with

$$\int_0^{+\infty} e^{-x}\sin(2x)\ \text{d}x = \dfrac{1}{2} - \dfrac{1}{4} \int_0^{+\infty} e^{-x}\sin(2x)\ \text{d}x $$

The two integrals are identical but opposite signs. Arrange and you will get:

$$\int_0^{+\infty} e^{-x}\sin(2x)\ \text{d}x = \dfrac{2}{5}$$

Enrico M.
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Since the integral consists of two transcendental functions $\sin(2x)$ and $e^{-x}$, you may use integration by parts. Set $u = \sin(2x)$ and $dv = e^{-x}dx$ and we get $$ \begin{align*} \int_{0}^{\infty} \sin(2x)e^{-x}dx &= \left[-\sin(2x)e^{-x} \right|_{0}^{\infty} + 2\int_{0}^{\infty}\cos(2x)e^{-x}dx \\ &= 2\int_{0}^{\infty}\cos(2x)e^{-x}dx \end{align*} $$ Doing integration by parts again with $u = \cos(2x)$ and $dv = e^{-x}dx$ we have $$ \begin{align*} \int_{0}^{\infty} \cos(2x)e^{-x}dx &= \left[-\cos(2x)e^{-x} \right|_{0}^{\infty} - 2\int_{0}^{\infty}\sin(2x)e^{-x}dx \\ &= 1-2\int_{0}^{\infty}\sin(2x)e^{-x}dx. \end{align*} $$ Putting everything together we obtain $$\int_{0}^{\infty} \sin(2x)e^{-x}dx = 2 - 4 \int_{0}^{\infty} \sin(2x)e^{-x}dx$$ Setting $y = \int_{0}^{\infty} \sin(2x)e^{-x}dx$ we then need to solve $$y = 2 - 4y \Rightarrow 5y = 2 \Rightarrow y = \int_{0}^{\infty} \sin(2x)e^{-x}dx = 2/5.$$

Daniel Flores
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