Page 2 of 2 FirstFirst 12
Results 16 to 20 of 20

Math Help - integral challenge

  1. #16
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Mathstud28 View Post
    You mean \frac{\pi}{2e} and yes....but I sense some Complex Analysis in the mix here

    EDIT: You corrected yourself.
    your integral = 2\int_0^{\infty} \frac{x \sin x}{x^2 + 1} \ dx - 2\int_0^{\infty} \frac{x \sin x}{(x^2 + 1)^2} \ dx, and by parts in the second integral gives us: your integral = \int_0^{\infty} \frac{x \sin x}{x^2 + 1} \ dx=\int_0^{\infty} \frac{\cos x}{x^2 + 1} \ dx why?

    and i guess i don't have to prove this very well-known fact that \int_0^{\infty} \frac{\cos x}{x^2 + 1} \ dx = \frac{\pi}{2e}.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I may as well finish how I done it. Overkill, yes. I wish I would have saw that Frullani thing right off. I am familiar with it.It did not occur to me. I will remember to look around for it from now on.

    By letting f(x)=\frac{e^{x}}{e^{x}+1}+C

    I left f(ax)-f(bx)=\frac{e^{ax}-e^{bx}}{(e^{ax}+1)(e^{bx}+1)}dx

    Then setting I(u)=\int_{0}^{u}\frac{f(ax)}{x}dx-\int_{0}^{u}\frac{f(bx)}{x}dx

    This is OK if \lim_{v\to 0}\frac{f(v)}{v} exists. Which works

    if C = -1/2.

    With C = -1/2, we get:

    I(u)=\int_{0}^{au}\frac{f(v)}{v}dv-\int_{0}^{bu}\frac{f(v)}{v}dv

    =\int_{au}^{bu}\frac{f(v)}{v}dv


    To save typng, I went on and eventually wound up with said solution.

    I like all of the methods. That is why I like to post an occasional integral.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by NonCommAlg View Post
    your integral = \int_0^{\infty} \frac{x \sin x}{x^2 + 1} \ dx=\int_0^{\infty} \frac{\cos x}{x^2 + 1} \ dx why?
    I assume the "why" was meant to be answere by me?

    Answer: Since \int_0^{\infty}\frac{x\sin(x)}{x^2+1} is even we have that \int_0^{\infty}\frac{x\sin(x)}{x^2+1}=\frac{1}{2}\  int_{-\infty}^{\infty}\frac{x\sin(x)}{x^2+1}

    Integrating by parts letting dv=\sin(x) and u=\frac{x}{x^2+1} gives

    \frac{1}{2}\int_{-\infty}^{\infty}\frac{x\sin(x)}{x^2+1}= \frac{1}{2}\left\{-\cos(x)\left(\frac{2}{x^2+1}-\frac{1}{x^2+1}\right)\right\}\bigg|_{x=-\infty}^{x=\infty}-\frac{1}{2}\int_{-\infty}^{\infty}\left\{\frac{-2\cos(x)}{(x^2+1)^2}+\frac{\cos(x)}{x^2+1}\right\}  dx

    Now \int_{-\infty}^{\infty}\frac{\cos(x)}{(x^2+1)^2}=\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}dx Why?

    So we get

    \begin{aligned}\frac{-1}{2}\int_{-\infty}^{\infty}\left\{\frac{-2\cos(x)}{(x^2+1)^2}+\frac{\cos(x)}{x^2+1}\right\}  &=\int_{-\infty}^{\infty}\frac{\cos(x)}{(x^2+1)^2}-\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}\\<br />
&=\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}-\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}dx\\<br />
&=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+1}\\<br />
&=\int_0^{\infty}\frac{\cos(x)}{x^2+1}dx\end{align  ed}

    Follow Math Help Forum on Facebook and Google+

  4. #19
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    11
    As for the well-known fact, see

    http://www.mathhelpforum.com/math-he...tegration.html

    last post.
    Follow Math Help Forum on Facebook and Google+

  5. #20
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Krizalid View Post
    As for the well-known fact, see

    http://www.mathhelpforum.com/math-he...tegration.html

    last post.
    Ahh that is one of my favorite ways to solve it...it is obviously much easier to use CA though...Ill be sure to use it when I learn it
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Challenge Integral
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: June 5th 2011, 04:39 AM
  2. Yet another integral challenge question
    Posted in the Math Challenge Problems Forum
    Replies: 8
    Last Post: May 11th 2010, 08:03 PM
  3. Challenge. (1)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 6th 2010, 03:37 PM
  4. Challenge integral
    Posted in the Calculus Forum
    Replies: 6
    Last Post: August 4th 2009, 12:52 PM
  5. Challenge
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 16th 2008, 08:13 AM

Search Tags


/mathhelpforum @mathhelpforum