How Can I Correctly Integrate e^(ix)cos(x)?

[flux!]

I'am trying to prove

\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}

Wolfram tells so http://integrals.wolfram.com/index.jsp?expr=e^(i*x)cos(x)&random=false

But I am stuck in obtaining the first term:

My step typically involved integration by parts:

let u=e^{ix}cos(x) and dv=dx

so:

du=-e^{ix}sin(x)dx+icos(x)e^{ix}du=ie^{ix}(sin(x)+cos(x))dxdu=ie^{2ix}dx

the other one is just: v=x

then:
\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\int xe^{2ix}dx

We will again do another integration by parts for the second term, so we let u=x and dv=e^{2ix}dx then solving further, we obtain:
\int xe^{2ix}dx=\frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}
plugging it back to the original problem, then doing simple distribution, we will obtain:
\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\left ( \frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}\right )
\int e^{ix}cos(x) dx=xe^{ix}cos(x)- \frac{1}{2}xe^{2ix}-\frac{1}{4}ie^{2ix}
notice that we have proved the 2nd term, but the other half is badly away from what the Integration Table and Tools says:
\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}

What have I gone wrong?

 

[SteamKing]

I'am trying to prove

\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}

Wolfram tells so http://integrals.wolfram.com/index.jsp?expr=e^(i*x)cos(x)&random=false

But I am stuck in obtaining the first term:

My step typically involved integration by parts:

let u=e^{ix}cos(x) and dv=dx

so:

du=-e^{ix}sin(x)dx+icos(x)e^{ix}du=ie^{ix}(sin(x)+cos(x))dxdu=ie^{2ix}dx

the other one is just: v=x

then:
\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\int xe^{2ix}dx

We will again do another integration by parts for the second term, so we let u=x and dv=e^{2ix}dx then solving further, we obtain:
\int xe^{2ix}dx=\frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}
plugging it back to the original problem, then doing simple distribution, we will obtain:
\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\left ( \frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}\right )
\int e^{ix}cos(x) dx=xe^{ix}cos(x)- \frac{1}{2}xe^{2ix}-\frac{1}{4}ie^{2ix}
notice that we have proved the 2nd term, but the other half is badly away from what the Integration Table and Tools says:
\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}

What have I gone wrong?

You can side step a lot of the work above if you remember:

$$cos\, x = \frac{e^{ix}+e^{-ix}}{2}$$

 

[beamie564]

I'm sorry, I couldn't follow your answer. But the integral is quite easy. Try using Euler's formula for e^x. Once you do ( and use a simple trigonometric identity) it's an easy
integration.
.
.
Edit: Or you can use the way in the post above mine, your convenience.

 

[Mark44]

I'am trying to prove

\int e^{ix}cos(x) dx= \frac{1}{2}x-\frac{1}{4}ie^{2ix}

All you really need to do is to show that $\frac d{dx}[\frac x 2 - \frac 1 4 ie^{2ix}] = e^{ix}\cos(x)$. This isn't that hard to do.

 

[mathman]

Use streamking's approach cos(x)=\frac{e^{ix}+e^{-ix}}{2}. It is now a trivial problem.

 

[flux!]

Darn! Applied steamking's approached, solved in 2 lines only. -_-

Thanks everyone!