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## Main Question or Discussion Point

I'am trying to prove

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

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 [itex]u=e^{ix}cos(x)[/itex] and [itex]dv=dx[/itex]

so:

[tex]du=-e^{ix}sin(x)dx+icos(x)e^{ix}[/tex][tex]du=ie^{ix}(sin(x)+cos(x))dx[/tex][tex]du=ie^{2ix}dx[/tex]

the other one is just: [itex]v=x[/itex]

then:

[tex]\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\int xe^{2ix}dx[/tex]

We will again do another integration by parts for the second term, so we let [itex]u=x[/itex] and [itex]dv=e^{2ix}dx[/itex] then solving further, we obtain:

[tex]\int xe^{2ix}dx=\frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}[/tex]

plugging it back to the original problem, then doing simple distribution, we will obtain:

[tex]\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\left ( \frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}\right )[/tex]

[tex]\int e^{ix}cos(x) dx=xe^{ix}cos(x)- \frac{1}{2}xe^{2ix}-\frac{1}{4}ie^{2ix}[/tex]

notice that we have proved the 2nd term, but the other half is badly away from what the Integration Table and Tools says:

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

What have I gone wrong?

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

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 [itex]u=e^{ix}cos(x)[/itex] and [itex]dv=dx[/itex]

so:

[tex]du=-e^{ix}sin(x)dx+icos(x)e^{ix}[/tex][tex]du=ie^{ix}(sin(x)+cos(x))dx[/tex][tex]du=ie^{2ix}dx[/tex]

the other one is just: [itex]v=x[/itex]

then:

[tex]\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\int xe^{2ix}dx[/tex]

We will again do another integration by parts for the second term, so we let [itex]u=x[/itex] and [itex]dv=e^{2ix}dx[/itex] then solving further, we obtain:

[tex]\int xe^{2ix}dx=\frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}[/tex]

plugging it back to the original problem, then doing simple distribution, we will obtain:

[tex]\int e^{ix}cos(x) dx=xe^{ix}cos(x)-i\left ( \frac{-i}{2}xe^{2ix}+\frac{1}{4}e^{2ix}\right )[/tex]

[tex]\int e^{ix}cos(x) dx=xe^{ix}cos(x)- \frac{1}{2}xe^{2ix}-\frac{1}{4}ie^{2ix}[/tex]

notice that we have proved the 2nd term, but the other half is badly away from what the Integration Table and Tools says:

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

What have I gone wrong?