Integral from 0 to infinite for the function cos(x)*e^-ax dx

[JonasBergby]

I am having some problems trying to solve this. I am supposed to decide an a >0 to give the integral the largest possible value. Then finding the value. The knowledge of how to do this seems to have slipped my mind entirely. Any takers?

I belive it is supposed to be 1, but i am not sure, nor do i know how to show that in a terms of calculations.

 

[skeeter]

Two iterations of integration by parts results in ...

[MATH]\lim_{b \to \infty} \bigg[\dfrac{e^{-ax}(\sin{x} - a \cos{x})}{a^2+1} \bigg]_0^b = 0 - \dfrac{-a}{a^2+1} = \dfrac{a}{a^2+1}[/math]
I would take the derivative of [MATH]\dfrac{a}{a^2+1}[/MATH] with respect to [MATH]a[/MATH] and determine the value of [MATH]a>0[/MATH] that yields a maximum using the first derivative test.

 

[Steven G]

What you wrote is [MATH]\int cos(x)*e^{-a}xdx[/MATH]. Is this correct?

In any case you solve the integral which will be a function of a. You want the max, so you have to find the critical point of this function of a. Lets see what you can do with this hint.

 

[JonasBergby]

Wow, thanks a lot. This is quite the helping hand. Apologies for writing it wrong, skeeter interpereted it right. Thanks again.

 

[Eduleo]

This kind of integrals, alternatively, can be solved using complex numbers:

[math]\displaystyle \int_{0}^{\infty} cos(x)*e^{-ax}dx = Real\left[ \int_{0}^{\infty} e^{-ax}e^{ix}dx \right]=Real\left[ \int_{0}^{\infty} e^{(-a+i)x}dx \right]=[/math]
[math]\displaystyle Real\left[\dfrac{1}{-a+i}e^{(-a+i)x}\right]_{0}^{\infty} = Real\left[-\dfrac{1}{a^2+1}e^{-ax}\left(\left( a\cos(x)-\sin(x)\right)+ (···)i\right)\right]_{0}^{\infty}= [/math]
[math]\left[\dfrac{1}{a^2+1}e^{-ax}\left(\sin(x)-a\cos(x)\right)\right]_{0}^{\infty}=\dfrac{a}{a^2+1}[/math]