Elementary exponential integral

  • Thread starter Master J
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  • #1
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Main Question or Discussion Point

I'm sure this integral is easy, but could someone perhaps show the working of:

[tex]\int[/tex] e[tex]^{i2t}[/tex] dt between t and 0.

I've tried it with trigonometric identities and keep getting lost!

Cheers!
 

Answers and Replies

  • #2
Char. Limit
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Try a u-substitution.

u=2it
du=2i dt
 
  • #3
AlephZero
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Why do you need to do a substitution?

It is exactly the same as

[tex]\int_0^t e^{at}\,dt[/tex]

except that [tex]a[/tex] is a complex number.
 
  • #4
Char. Limit
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Although strictly speaking, it's bad form to use the same variable in your bounds and your integrand.
 
  • #5
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Cheers guys.

Yea true I guess, perhaps it should have been tau as the differential.
 
  • #6
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This relates to the integral of exp[ iwt], between t and 0. This is then squared, so i was going to just integrate exp [ 2iwt].


The answer is 2(1 - coswt) / w ....i cant seem to get this at all. Any ideas?
 
  • #7
Char. Limit
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Is it the integral that's squared, or just the integrand?
 
  • #8
HallsofIvy
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The integral
[tex]\int_0^t e^{-2i\tau}d\tau[/tex]
is
[tex]-\fra{1}{2i}e^{-2it}= \frac{i}{2}e^{-2it}[/tex]

NOT
[tex]\frac{2(1- cos(\omega t))}{\omega}[/tex]
or even
[tex]\frac{2(1- cos(2t))}{2}= 1- cos(2t)[/itex]
 

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