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Math Help - Improper Integrals

  1. #1
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    Improper Integrals

    The question is read
    integrate---

    int(0 to infinity) (e^-x)(sinx)

    So i first expressed this as a limit where

    limit A approaching infinity (int 0 to A)(e^-x)(sinx).

    I integrated by parts twice to find that int of (e^-x)(sinx) = (e^-x (-sinx-cosx))/2

    But now i am stuck on solving the limit itself. Any help is appreciated Thanks in advance
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by dreamx20 View Post
    The question is read
    integrate---

    int(0 to infinity) (e^-x)(sinx)

    So i first expressed this as a limit where

    limit A approaching infinity (int 0 to A)(e^-x)(sinx).

    I integrated by parts twice to find that int of (e^-x)(sinx) = (e^-x (-sinx-cosx))/2

    But now i am stuck on solving the limit itself. Any help is appreciated Thanks in advance
    for integrating by parts how did u chose your  u and  dv ???

     \displaystyle \int _0 ^{\infty} e^{-x} \sin {x} \; dx = \frac {1}{2}


    P.S. u can use formula :

     \displaystyle \int e^{(\alpha x)} \sin {(\beta x)} \; dx = \frac {e^{(\alpha x)} (-\beta \cos(\beta x) +\alpha \sin {(\beta x)})}{\alpha^2 + \beta^2}=-\frac{1}{2} e^{-x} (\sin{x} +\cos {x}) +C

    and just put limits
    Last edited by yeKciM; August 10th 2010 at 09:52 AM.
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  3. #3
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    sorry the actual question read int (from 0 to infinity) 2(e^-x)sin(x) which was integrated by parts again twice..cause we did not learn the identity for the course. where u=sinx and v=e^-x...

    I still havent figured this question out and its driving me nuts...please help
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by dreamx20 View Post
    sorry the actual question read int (from 0 to infinity) 2(e^-x)sin(x) which was integrated by parts again twice..cause we did not learn the identity for the course. where u=sinx and v=e^-x...

    I still havent figured this question out and its driving me nuts...please help
    That is the method of "by parts" which you can readily google. There is nothing complicated with that approach; simply select the appropriate U' and dV' and be on your way. Here is another approach known as "complexifying the integral"

     2 \int_0^{ \infty } e^{ -x } sin(x) dx

     2 \int_0^{ \infty } Re [  e^{ -x } ( i cosx - sinx ) ]  dx

     2 \int_0^{ \infty } Re [  e^{ -x } ( ie^{- ix} ) ]  dx

     2 \int_0^{ \infty } Re [ ( ie^{- ix - x} ) ]  dx

     2 Re ( \int_0^{ \infty } ie^{-x( i + 1 )}    dx )

     2 Re ( - i \frac{ e^{ -x(i+1) }}{1+i} )

     2 Re ( - i \frac{ e^{ -x(i+1) }}{1+i} * \frac{1-i}{1-i})

     2 Re ( - i \frac{ (1-i) e^{ -x(i+1) }}{2} )

     2 Re ( - i \frac{ (1-i) }{2} e^{-x} e^{-xi} )

     2 Re ( - \frac{ (1+i) }{2} e^{-x} e^{-xi} )

     2 Re ( - \frac{ (1+i) }{2} e^{-x} [ cosx - isinx ] )

     2 Re ( - e^{-x} \frac{cosx-isinx}{2} - i e^{-x} \frac{cosx-isinx}{2} )

     -( e^{-x} cosx + e^{-x} sinx )

    Evaluate the above at the given bounds to find the answer
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  5. #5
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    The way to do it by parts is this


    [LaTeX ERROR: Convert failed]
    [LaTeX ERROR: Convert failed] (here I've used your choice for [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] )
    [LaTeX ERROR: Convert failed] (here I've used,  u = e^{-x} and dv = \cos(x)\; dx)

    This gives us

    [LaTeX ERROR: Convert failed]

    Taking the limit as x approaches \infty, we get 0 because the exponential goes to 0 while the trig functions keep cycling between -1 and 1.
    Plugging in 0 gives us -(0+1) = -1

    So in the end the integral should be 0-(-1) = 1
    Last edited by Vlasev; August 10th 2010 at 06:49 PM.
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  6. #6
    MHF Contributor chisigma's Avatar
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    In any table of Laplace Tranforms You find...

    \displaystyle \mathcal{L} \{\sin t \} = \int_{0}^{\infty} e^{- s t}\ \sin t\ dt = \frac{1}{1+s^{2}} , \Re (s) > 0 (1)

    ... so that setting in (1) s=1 You have...

    \displaystyle \int_{0}^{\infty} e^{- t}\ \sin t\ dt = \frac{1}{2} (2)

    Kind regards

    \chi \sigma
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