Improper Integrals

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August 10th 2010, 09:34 AM
dreamx20

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
August 10th 2010, 09:38 AM
yeKciM

Quote:

Originally Posted by dreamx20

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 :D
August 10th 2010, 06:03 PM
dreamx20

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
August 10th 2010, 06:27 PM
AllanCuz

Quote:

Originally Posted by dreamx20

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 :D
August 10th 2010, 06:39 PM
Vlasev

The way to do it by parts is this

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[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

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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
August 10th 2010, 08:58 PM
chisigma

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

$\chi$ $\sigma$