Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that you can integrate

$$\int e^{-x}\cos(x)dx$$

by parts, but I would like to know how you can use complex variables instead.

share|cite|improve this question
Use Euler's formula to express $\cos x$ in terms of complex exponentials. – Ben Crowell Feb 5 '12 at 4:24
Re integration by parts -- is that possible for this integral? I don't see how that would work. – Ben Crowell Feb 5 '12 at 4:54
@Ben I think it is possible to integrate this by parts. In general it is possible to integrate $\int e^{ax} \cos bx \mathrm d x$ by parts. The same is true when $\cos bx$ is replaced by $\sin bx$. – user21436 Feb 5 '12 at 5:11

Euler's form of a complex Number:

$$e^{i x}=\cos x+i \sin x$$

And, note that, as $\sin (-x)= -\sin x$ and $\cos (-x) = \cos x$, we have that, $$e^{-i x}=\cos x-i \sin x$$

This together gives you, $$\cos x = \dfrac{e^{ix}+e^{-ix}}{2}$$

Note: $\Re$ stands for the real part of a complex number and $\Im$ for its imaginary part.

Method 1:

$$\begin{align*}\int e^{-x}\cos x~~ \mathrm d x &= \Re\left(\int e^{-x}e^{ix} \mathrm dx\right)\\&=\Re\left(\int e^{x(i-1)} \mathrm d x\right) \\&=\Re\left(\dfrac{e^{x(i-1)}}{i-1}\right)\\&=e^{-x}\cdot\Re\left(\dfrac{\cos x+ i \sin x}{-1+i}\right) \\&= e^{-x} \cdot\dfrac{\sin x-\cos x}{2}\end{align*} $$

Method 2:(Ben's Comment)

$$\begin{align*} \int e^{-x} \cos x ~\mathrm dx&=\int e^{-x} \cdot \dfrac{e^{ix}+e^{-ix}}{2} \mathrm dx\\&=\dfrac{1}{2} \cdot \int (e^{ix-x}+e^{-ix-x}) \mathrm d x\\&=\dfrac{1}{2}\cdot \left(\dfrac{e^{x(i-1)}}{i-1}+\dfrac{e^{x(-1-i)}}{-i-1}\right)\end{align*}$$

Leaving the simplifications to you, you will see that the answer still turns out to be the same.

share|cite|improve this answer
Looks fine, except that the factor of 1/2 appears one line too early. As an alternative to using the real part, one can write $\cos x$ as $(e^{ix}+e^{-ix})/2$. – Ben Crowell Feb 5 '12 at 4:51
@Ben I have added your suggestion as well. Thank You! – user21436 Feb 5 '12 at 5:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.