Complexifying an Integral: How to Solve Tricky Integrals Using Complex Numbers

[jellicorse]

Watching a video on differential equations by Arthur Mattuck of MIT, I came across a method which was new to me for solving certain integrals, such as \int e^{-x}cos(x)dx . part of the video is here:

Given that e^{ix}=cosx+isinx, this integral can be re-written as Re\int e^{ix}\cdot e^{-x}dx = Re\int e^{x(-1+i)}dx and integrated this way, avoiding the need for reduction formulas...

I have looked up a few examples of this and think I can understand it when cos(x) is part of the integral, but am unsure when it comes to sin(x).

There is a (now closed) thread on this forum https://www.physicsforums.com/showthread.php?t=511534&highlight=complexifying+integral which I have been attempting to follow but I can't see quite how it works. I was wondering if anyone could help me to understand this.

The following is effectively the working from that post:

\int e^x sin2x dx

=Re\int e^x \cdot e^{i(\frac{\pi}{2}-2x)}dx

I can see that since the real part of polar co-ordinates is contained in the "cos(2x)" part, they have put the Sin(2x) in terms of Cos(2x) by subtracting it from \frac{\pi}{2}.

But I don't see how the next step is reached:

=Re\frac{i}{1-2i}e^{(1-2i)x}


I'm not sure how they got to this, nor what the intermediate steps might be. If I was doing this, I would have gone along the lines of Re\int e^{x+i(\frac{\pi}{2}-2x)}dx. While I can see that that expression doesn't look very attractive to try to integrate, I can not see how the author got to e^{(1-2)x}

 

[pasmith]

Watching a video on differential equations by Arthur Mattuck of MIT, I came across a method which was new to me for solving certain integrals, such as \int e^{-x}cos(x)dx . part of the video is here:

Given that e^{ix}=cosx+isinx, this integral can be re-written as Re\int e^{ix}\cdot e^{-x}dx = Re\int e^{x(-1+i)}dx and integrated this way, avoiding the need for reduction formulas...

I have looked up a few examples of this and think I can understand it when cos(x) is part of the integral, but am unsure when it comes to sin(x).

If you want sin(x), take the imaginary part of e^{ix} instead of the real part.

 

[jellicorse]

OK, thanks Pasmith...