Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Prove $\exists \xi \in [0,\pi]$ such that $$ \int_0^\pi e^{-x}\cos(x)\,dx = \sin(\xi)$$

Well, I was thinking on using the Generalized MVT, but it doesnt seem I obtain the right answer. How do you guys would tackle this problem?

share|cite|improve this question
up vote 6 down vote accepted

$$\left|\int_0^\pi\mathrm e^{-x}\cos(x)\,\mathrm dx\right|\lt\int_0^\pi\mathrm e^{-x}\,\mathrm dx=\left[-\mathrm e^{-x}\right]_0^\pi=1-\mathrm e^{-\pi}\lt1\;.$$

share|cite|improve this answer
Actually if $\xi$ is supposed to be in $[0,\pi]$, you also want it to be $\ge 0$. For that, apply the change of variables $u =\pi - x$ to obtain $$\int_{\pi/2}^\pi e^{-x} \cos(x)\ dx = -\int_0^{\pi/2} e^{x-\pi} \cos(x)\ dx$$ so $$ \int_0^\pi e^{-x} \cos(x)\ dx = \int_0^{\pi/2} (e^{-x} - e^{x-\pi}) \cos(x)\ dx \ge 0$$ since $e^{-x} - e^{x-\pi}$ for $0 \le x \le \pi/2$. – Robert Israel Oct 20 '12 at 0:02

Since you want to prove existence, one example will suffice.

Note that $$\int_0^\pi {{e^{ - x}}\cos (x)dx} = \frac{{{e^{ - \pi }}(1 + {e^\pi })}}{2}$$ (you can use complex exponential trick to evaluate the integral) and let $$\xi = {\sin ^{ - 1}}\left( {\frac{{{e^{ - \pi }}(1 + {e^\pi })}}{2}} \right) \in \left[ {0,\pi } \right].$$

share|cite|improve this answer

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.