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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 \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}
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}
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