Solve Integral Equation: xe-axcos(x)dx from 0 to ∞

[WWCY]

Homework Statement

Solve from x = 0 to x = ∞, ∫xe^-axcos(x)dx

Homework Equations

The Attempt at a Solution

I have a solution for the integral ∫e^-axcos(x)dx at the same limits, and I feel that the result might be of use, but have no idea how to manipulate the integral above such that I can use the result.

Here's what I tried so far,

Definite Integral by parts into

[x∫e^-axcos(x)dx] - ∫ (∫e^-axcos(x)dx) dx,

left hand term reduces to x(sinx - acosx) / [ e^ax(a² + 1) ]

while the right term seems something that I can work with with some effort, the left is proving problematic with the x heading towards infinity. I felt that I needed to somehow prove that the left term tends to zero at infinity, and that I could try some nasty squeeze theorems on the trigo functions. However, this method doesn't seem to require the use of the result of the simpler integral; it doesn't feel quite right/efficient.

Assistance is appreciated, thank you.

 

[MarkFL]

We are given to evaluate:

$\displaystyle I=\int_0^{\infty}xe^{-ax}\cos(x)\,dx=\lim_{t\to\infty}\left(\int_0^{t}xe^{-ax}\cos(x)\,dx\right)$

If you already know:

$\displaystyle I_1=\int e^{-ax}\cos(x)\,dx$

Then you may use IBP, where:

$u=x\implies du=dx$

$dv=e^{-ax}\cos(x)\,dx\implies v=I_1$

What do you get when you substitute for $I_1$?

 

[WWCY]

We are given to evaluate:

What do you get when you substitute for $I_1$?

Thanks for the reply,

May I ask if $I_1$ is referring to a definite or indefinite integral?

 

[MarkFL]

Thanks for the reply,

May I ask if $I_1$ is referring to a definite or indefinite integral?

I intended it to refer to the indefinite integral.

 

[WWCY]

I intended it to refer to the indefinite integral.

I have the expression:

(sinx - acosx) / [ eax(a2 + 1) ]

Is this what you meant?

 

[MarkFL]

Okay, so knowing (up to the constant of integration) that:

$I_1=\dfrac{\sin(x)-a\cos(x)}{e^{ax}\left(a^2+1\right)}$

Then what does our expression for $I$ become when applying IBP?

 

[Buffu]

Use $\cos x = \dfrac{e^{ix} + e^{-ix}}{2}$.

 

[Mark44]

@WWCY, in the future, please post questions about integrals (and derivatives) in the Calculus & Beyond section, not the Precalc section.

 

[Ray Vickson]

Homework Statement

Solve from x = 0 to x = ∞, ∫xe^-axcos(x)dx

Homework Equations

The Attempt at a Solution

I have a solution for the integral ∫e^-axcos(x)dx at the same limits, and I feel that the result might be of use, but have no idea how to manipulate the integral above such that I can use the result.

Here's what I tried so far,

Definite Integral by parts into

[x∫e^-axcos(x)dx] - ∫ (∫e^-axcos(x)dx) dx,

left hand term reduces to x(sinx - acosx) / [ e^ax(a² + 1) ]

while the right term seems something that I can work with with some effort, the left is proving problematic with the x heading towards infinity. I felt that I needed to somehow prove that the left term tends to zero at infinity, and that I could try some nasty squeeze theorems on the trigo functions. However, this method doesn't seem to require the use of the result of the simpler integral; it doesn't feel quite right/efficient.

Assistance is appreciated, thank you.

I am surprised that you have not met one of the standard tricks used for such problems. If we let $F(a) = \int_0^{\infty} e^{-ax} \cos(x) \, dx$, then---if we can differentiate inside the integral sign--- we have
$$ F^{\prime}(a) = \int_0^{\infty} \cos(x) \frac{\partial e^{-ax}}{\partial a} \, dx = - \int_0^{\infty} x e^{-ax} \cos(x) \, dx.$$
There are theorems available that justify interchanging $\partial / \partial a$ and $\int$; you can look them up.

 

[Ray Vickson]

Homework Statement

Solve from x = 0 to x = ∞, ∫xe^-axcos(x)dx

Homework Equations

The Attempt at a Solution

I have a solution for the integral ∫e^-axcos(x)dx at the same limits, and I feel that the result might be of use, but have no idea how to manipulate the integral above such that I can use the result.

Here's what I tried so far,

Definite Integral by parts into

[x∫e^-axcos(x)dx] - ∫ (∫e^-axcos(x)dx) dx,

left hand term reduces to x(sinx - acosx) / [ e^ax(a² + 1) ]

while the right term seems something that I can work with with some effort, the left is proving problematic with the x heading towards infinity. I felt that I needed to somehow prove that the left term tends to zero at infinity, and that I could try some nasty squeeze theorems on the trigo functions. However, this method doesn't seem to require the use of the result of the simpler integral; it doesn't feel quite right/efficient.

Assistance is appreciated, thank you.

So, you want to evaluate
$$\lim_{U \to \infty} \left. \frac{x ( \sin x - a \cos x) e^{-ax}}{a^2+1} \right|_0^{U}\hspace{2ex}?$$
For $a > 0$, what is $\lim_{U \to \infty} F(U) e^{-aU}$ for any bounded function $F(U)$? For $a >0$ what is
$\lim_{U \to \infty} U^n F(U) e^{-aU}$ for any bounded function $F(U)$ and any finite $n > 0$?

 

[WWCY]

Hi everyone, I went to take a look differentiating under the integral sign, and found that to be so much more efficient in solving the problem. For good measure, I've also managed to sort it out using IBP.

Many thanks to everyone!

 

[Ray Vickson]

Hi everyone, I went to take a look differentiating under the integral sign, and found that to be so much more efficient in solving the problem. For good measure, I've also managed to sort it out using IBP.

Many thanks to everyone!

What does "IBP" mean?

 

[epenguin]

What does "IBP" mean?

I didn't dare ask.

 

[WWCY]

What does "IBP" mean?

Oops, I meant integration by parts.