Results 1 to 3 of 3

Thread: Interesting Integral

  1. #1
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Interesting Integral

    Show that for $\displaystyle a \ne 0, a \in \mathbb{R}$ that

    $\displaystyle \int_{0}^{\pi}\cos(a\sin(\theta))e^{a\cos(\theta)} d\theta=\pi$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by TheEmptySet View Post

    Show that for $\displaystyle a \ne 0, a \in \mathbb{R}$ that $\displaystyle \int_{0}^{\pi}\cos(a\sin(\theta))e^{a\cos(\theta)} d\theta=\pi$
    it's nice but probably not hard enough!

    Spoiler:


    $\displaystyle e^{ae^{i \theta}}=e^{a \cos \theta}e^{ia\sin \theta}=e^{a \cos \theta}(\cos(a \sin \theta) + i \sin(a \sin \theta)).$ therefore: $\displaystyle e^{a \cos \theta} \cos(a \sin \theta) = \text{Re}(e^{ae^{i \theta}})=\text{Re} \left(\sum_{n=0}^{\infty}\frac{a^ne^{in \theta}}{n!} \right)=\sum_{n=0}^{\infty} \frac{a^n \cos(n \theta)}{n!}.$

    hence $\displaystyle \int e^{a \cos \theta} \cos(a \sin \theta) \ d \theta = \theta + \sum_{n=1}^{\infty} \frac{a^n \sin(n \theta)}{n \cdot n!} +C.$ your definite integral is just a simple result of what we have now!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Wow that is a very nice solution. Better than this, but here is another way....

    Consider the integral where $\displaystyle \gamma$ is the unit circle

    $\displaystyle \int_\gamma \frac{e^{az}}{z}dz$ in the complex plane. Then by cauchy's integral formula or the residue theorem.

    $\displaystyle \int_\gamma \frac{e^{az}}{z}dz=2\pi i$

    Parameterize the curve with $\displaystyle z=e^{i\theta}$ for $\displaystyle -\pi \le \theta \le \pi$ and then $\displaystyle dz=ie^{i\theta}d\theta=izd\theta$=

    $\displaystyle \int_{-\pi}^{\pi} \frac{e^{ae^{i\theta}}}{z}izd\theta=i\int_{-\pi}^{\pi}e^{a\cos(\theta) +ia\sin(\theta)}d\theta=i\int_{-\pi}^{\pi}e^{a\cos(\theta)}[\cos(a\sin(\theta)+i\sin(a\sin(\theta))]d\theta=$

    $\displaystyle i\int_{-\pi}^{\pi}\cos(a\sin(\theta)e^{a\cos(\theta)}d\the ta-\int_{-\pi}^{\pi}\sin(a\sin(\theta))e^{a\cos(\theta)}d\th eta$

    So we know that this equals $\displaystyle 2\pi i$ setting immaginary parts equal we get

    $\displaystyle 2\pi =\int_{-\pi}^{\pi}\cos(a\sin(\theta)e^{a\cos(\theta)}d\the ta$

    Since this is an even function we get (Finally)

    $\displaystyle \pi =\int_{0}^{\pi}\cos(a\sin(\theta)e^{a\cos(\theta)} d\theta$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Interesting Integral
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Jun 16th 2011, 03:08 AM
  2. Interesting integral
    Posted in the Math Challenge Problems Forum
    Replies: 7
    Last Post: Aug 27th 2009, 12:29 PM
  3. An interesting integral
    Posted in the Math Challenge Problems Forum
    Replies: 5
    Last Post: Jul 3rd 2009, 10:35 PM
  4. Another interesting integral
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: Jun 8th 2009, 02:03 AM
  5. An interesting integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 17th 2009, 07:42 AM

Search Tags


/mathhelpforum @mathhelpforum