# Improper Integrals

• Aug 10th 2010, 09:34 AM
dreamx20
Improper Integrals
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
• Aug 10th 2010, 09:38 AM
yeKciM
Quote:

Originally Posted by dreamx20
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 $\displaystyle u$ and $\displaystyle dv$ ???

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

P.S. u can use formula :

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

and just put limits :D
• Aug 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...

• Aug 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...

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"

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle -( e^{-x} cosx + e^{-x} sinx )$

Evaluate the above at the given bounds to find the answer :D
• Aug 10th 2010, 06:39 PM
Vlasev
The way to do it by parts is this

$\displaystyle \displaystyle \int 2 e^{-x} \sin {x} \; dx =$
$\displaystyle \displaystyle = -2e^{-x}\sin(x)+2\int e^{-x}\cos(x)\; dx \;\;$ (here I've used your choice for $\displaystyle u$ and $\displaystyle dv$)
$\displaystyle \displaystyle = -2e^{-x}\sin(x) - 2e^{-x}\cos(x) - 2\int e^{-x} \sin {x} \; dx \;\;$ (here I've used, $\displaystyle u = e^{-x}$ and $\displaystyle dv = \cos(x)\; dx$)

This gives us

$\displaystyle \displaystyle \int 2 e^{-x} \sin {x} = -e^{-x}(\sin(x)+\cos(x))$

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

So in the end the integral should be 0-(-1) = 1
• Aug 10th 2010, 08:58 PM
chisigma
In any table of Laplace Tranforms You find...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$