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

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In summary, Arthur Mattuck of MIT explains how to solve certain integrals using a new method which avoids the need for reduction formulas. This new method is based on the real and imaginary parts of the polar co-ordinates, and can be applied to sin(x) as well as cos(x).
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jellicorse
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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 [tex]\int e^{-x}cos(x)dx[/tex] . part of the video is here:

Given that [tex]e^{ix}=cosx+isinx[/tex], this integral can be re-written as [tex]Re\int e^{ix}\cdot e^{-x}dx[/tex] = [tex]Re\int e^{x(-1+i)}dx[/tex] 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:

[tex]\int e^x sin2x dx[/tex]

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

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 [tex]\frac{\pi}{2}[/tex].

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

[tex]=Re\frac{i}{1-2i}e^{(1-2i)x}[/tex]


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 [tex] Re\int e^{x+i(\frac{\pi}{2}-2x)}dx[/tex]. 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 [tex]e^{(1-2)x}[/tex]
 
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jellicorse said:
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 [tex]\int e^{-x}cos(x)dx[/tex] . part of the video is here:

Given that [tex]e^{ix}=cosx+isinx[/tex], this integral can be re-written as [tex]Re\int e^{ix}\cdot e^{-x}dx[/tex] = [tex]Re\int e^{x(-1+i)}dx[/tex] 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 [itex]e^{ix}[/itex] instead of the real part.
 
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OK, thanks Pasmith...
 

1. What does it mean to "complexify" an integral?

Complexifying an integral means to extend the domain of integration from real numbers to complex numbers. This allows for a wider range of functions to be integrated, as complex numbers have both real and imaginary components.

2. Why would someone want to complexify an integral?

Complexifying an integral can be useful in solving certain types of integrals, particularly those involving trigonometric functions or complex functions. It can also help in evaluating improper integrals that would otherwise be divergent.

3. How is an integral complexified?

To complexify an integral, we replace any real variables in the integrand with complex variables. For example, the integral of ex can be complexified by replacing x with z, where z is a complex number.

4. Are there any limitations to complexifying an integral?

Complexifying an integral does not guarantee a solution. It may make the integral more complex and difficult to solve. Additionally, the resulting solution may only be valid in certain regions of the complex plane.

5. Is there a difference between complexifying and integrating a complex function?

Yes, there is a difference. Complexifying an integral involves extending the domain of integration, while integrating a complex function involves finding the antiderivative of a complex-valued function. These two concepts are related, but not the same.

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