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Thread: Calc of Residues

  1. #1
    Feb 2009

    Calc of Residues

    Hi! I've learned how to take contour integrals for functions that require branch cuts to find improper integrals. However, is there a way to find, say, $\displaystyle \int_0^{\infty}\frac{\sin x}{\sqrt{x}}dx$ using a similar method? I haven't seen any examples involving trig functions with multifunctions.

    Thanks for any help!
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  2. #2
    Eater of Worlds
    galactus's Avatar
    Jul 2006
    Chaneysville, PA
    Yes, we can do this with CA. I am skiddish about doing this. Hope I don't make a mistake somewhere.

    $\displaystyle \int_{0}^{\infty}\frac{sin(x)}{\sqrt{x}}dx$

    Make the sub $\displaystyle u=\sqrt{x},\;\ u^{2}=x$

    Then, we can write it as $\displaystyle 2\int_{0}^{\infty}sin(u^{2})du$

    Try using a 'wedge' in the 1st quadrant of the complex plane with angle

    $\displaystyle {\phi}=\frac{\pi}{4}$

    Now, using $\displaystyle f(z)=e^{z^{2}}$

    By the residue theorem, $\displaystyle \int e^{-z^{2}}dz=2{\pi}i\sum\text{enclosed residues}$

    Because there are no singularities, the enclosed residue = 0.

    Since the wedge is broken up into three different sections along the contour, we can break it up into three separate integrals. Call them a,b, and c for instance.

    $\displaystyle \int e^{-z^{2}}dz=\int_{a} e^{-z^{2}}dz+\int_{b}e^{-z^{2}}dz+\int_{c}e^{-z^{2}}dz$

    Take the radius of the wedge to be fixed for now at r=R and use the polar:

    $\displaystyle z=re^{i{\theta}}$

    $\displaystyle dz=dr(e^{i{\theta}})+ire^{i{\theta}}$

    First, take a peek at curve a, which lies on the x-axis.

    Along this curve, $\displaystyle d{\theta}=0, \;\ {\theta}=0, \;\ z=r, \;\ dz=dr$

    Curve b, is circular in its path at radius R, so while $\displaystyle {\theta}$ varies, r is fixed. So, along b, $\displaystyle r=R, \;\ dr=0, \;\ dz=iRe^{i{\theta}}d{\theta}$.

    Next, along c, $\displaystyle {\theta}$ is fixed. But this time $\displaystyle {\theta}=\frac{\pi}{4}, \;\ d{\theta}=0, \;\ dz=e^{i\frac{\pi}{4}}dr$

    So, the integral from above becomes:

    $\displaystyle \int e^{-z^{2}}dz=\int_{0}^{R}e^{-z^{2}}dr+\int_{0}^{\frac{\pi}{4}}e^{-R^{2}e^{i2{\theta}}}iRe^{i{\theta}}d{\theta}+\int_ {0}^{R}e^{-r^{2}e^{i2{\theta}}}e^{\frac{i{\pi}}{4}}dr$

    Now, let $\displaystyle R\to {\infty}$. On b, as $\displaystyle R\to {\infty}$,

    we get $\displaystyle \lim_{R\to {\infty}}\int_{0}^{\frac{\pi}{4}}e^{-R^{2}e^{i2{\theta}}}iRe^{i{\theta}}d{\theta}=0$

    because $\displaystyle e^{-R^{2}e^{i2{\theta}}}\to 0$ is a lot faster than $\displaystyle R\to {\infty}$.

    Remember that $\displaystyle \int e^{-z^{2}}dz=0$. Then we are left with

    $\displaystyle \lim_{R\to {\infty}}\left[\int_{0}^{R}e^{-r^{2}}dr-\int_{0}^{R}e^{-r^{2}e^{2i{\theta}}}e^{\frac{{\pi}i}{4}}dr\right]=0$

    The first one is a familiar one which equals $\displaystyle \int_{0}^{\infty}e^{-r^{2}}dr=\frac{\sqrt{\pi}}{2}$

    By Ol' Euler: $\displaystyle e^{\frac{{\pi}i}{4}}=cos(\frac{\pi}{4})+isin(\frac {\pi}{4})=\frac{1+i}{\sqrt{2}}$

    $\displaystyle \frac{\sqrt{\pi}}{2}=\frac{1}{\sqrt{2}}\int_{0}^{\ infty}cos(r^{2})dr+\frac{1}{\sqrt{2}}\int_{0}^{\in fty}sin(r^{2})dr+\frac{i}{\sqrt{2}}\int_{0}^{\inft y}[cos(r^{2})dr-sin(r^{2})]dr$

    Equate real and imaginary and note that $\displaystyle Im(\frac{\sqrt{\pi}}{2})=0$

    $\displaystyle \frac{i}{\sqrt{2}}\int_{0}^{\infty}[cos(r^{2})-sin(r^{2})]dr=0$

    and $\displaystyle \int_{0}^{\infty}cos(r^{2})dr=\int_{0}^{\infty}sin (r^{2})dr$

    $\displaystyle \frac{\sqrt{\pi}}{2}=\frac{1}{\sqrt{2}}\int_{0}^{\ infty}cos(r^{2})dr+\frac{1}{\sqrt{2}}\int_{0}^{\in fty}sin(r^{2})dr=\frac{2}{\sqrt{2}}\int_{0}^{\inft y}sin(r^{2})dr$

    $\displaystyle \int_{0}^{\infty}sin(r^{2})dr=\frac{1}{2}\sqrt{\fr ac{\pi}{2}}$

    Therefore, from the top, $\displaystyle \int_{0}^{\infty}\frac{sin(x)}{\sqrt{x}}dx=\sqrt{\ frac{\pi}{2}}$
    Attached Thumbnails Attached Thumbnails Calc of Residues-contour.gif  
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  3. #3
    Super Member Random Variable's Avatar
    May 2009
    I'm probably going to regret asking this, but why does $\displaystyle e^{-r^{2}e^{2i \theta}}\Big(\frac{1+i}{\sqrt{2}}\Big) = \frac{1}{\sqrt{2}} \Big(\cos r^{2} + \sin r^{2}\Big) + \frac{i}{\sqrt{2}} \Big(\cos r^{2} - \sin r^{2} \Big) $ ?
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  4. #4
    Super Member
    Jan 2009
    Using these formulas ( obtained by Euler )

    $\displaystyle \int_0^{\infty} x^{n-1} e^{-px} \cos(qx) ~dx = \frac{ \Gamma(n) }{ r^n} \cos(n \alpha)$ and

    $\displaystyle \int_0^{\infty} x^{n-1} e^{-px} \sin(qx) ~dx = \frac{ \Gamma(n) }{ r^n} \sin(n \alpha) $

    $\displaystyle r = \sqrt{ p^2 + q^2 } ~~~ \alpha = \tan^{-1}( \frac{ q}{p} ) $
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