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Thread: integral challenge

  1. #16
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    Quote Originally Posted by Mathstud28 View Post
    You mean $\displaystyle \frac{\pi}{2e}$ and yes....but I sense some Complex Analysis in the mix here

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

    and i guess i don't have to prove this very well-known fact that $\displaystyle \int_0^{\infty} \frac{\cos x}{x^2 + 1} \ dx = \frac{\pi}{2e}.$
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  2. #17
    Eater of Worlds
    galactus's Avatar
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    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 $\displaystyle f(x)=\frac{e^{x}}{e^{x}+1}+C$

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

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

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

    if C = -1/2.

    With C = -1/2, we get:

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

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


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

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

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

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

    So we get

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

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  4. #19
    Math Engineering Student
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    As for the well-known fact, see

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

    last post.
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  5. #20
    MHF Contributor Mathstud28's Avatar
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    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
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