Solve Exponential Integral - Get Help Now

In summary, the conversation discusses a request for help with a simple integration involving the integral of exp(iab/c)exp(-iaz) from 0 to infinity. Some participants suggest various approaches and solutions, but there is also a debate about the conditions for convergence, with some pointing out potential mistakes in previous responses. Ultimately, it is concluded that the condition for convergence is when the real part of i((b/c)-z) is less than 0.
  • #1
Skullmonkee
22
0
Hi i was wondering if anyone could help me with a fairly simple integration.

[tex]\int exp(iab/c)exp(-iaz) da[/tex]

Where the integral is from 0 to infinity

Im not sure how to proceed with this and am basically teaching myself
Thanks in advance for any help.
 
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  • #2
Do you have any conditions on b,c, and z? This integral is not going to be solvable as far as I can tell. Remember, [tex] e^{ix} = cos(x) + isin(x)[/tex] and you'll end up trying to take a limit at infinity which does not exist for cosines and sines.
 
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  • #3
I say the answer is 0
First consolidate the integrand to exp(iaK) where K = (b/c) - z, then expand the exponential to cos(aK) +i*sin(aK)
then [tex]
\int cos(aK) da + i\int sin(aK) da = 0
[/tex]
since integrating a sine or cosine gives zero, since there are equal positive and negative areas for every cycle.
 
  • #4
ACPower said:
since integrating a sine or cosine gives zero, since there are equal positive and negative areas for every cycle.

unless the limit is at infinity, which means the limit is not defined.
 
  • #5
In the wording of the problem, Skullmonkee forgot to specify if the coefficients b, c, z are real or complex.
Suppose b, c real and z complex. If z=x+i y , with x , y real and y<0
then, the integral (from 0 to infinity) is convergent = 1/(-y+i(x-b))
 
  • #6
The correct answer follows:

Let [itex]k = i(b/c-z)[/itex]. Then the integral is the same as
[tex]\int_0^\infty \exp(ka) \;da[/tex]
which converges precisely when [itex]\operatorname{Re}(k) < 0[/itex]. In this case, the value of the integral is
[tex]\frac{1}{k} \exp(ka) \biggr|_{a=0}^\infty = -\frac{1}{k} = \frac{i}{b/c-z}.[/tex]

Pengwuino and ACPower assume that [itex]b/c - z[/itex] is real, but if that is not the case, the integrand may not be purely sinusoidal (that is, [itex]k[/itex] above may have nonzero real part). JJacquelin gives an incorrect value for the integral (there is a missing c).
 
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  • #7
JJacquelin gives an incorrect value for the integral (there is a missing c), as well as incorrect conditions for convergence.
Adriank is right, I forgot c in my equation. Thanks for bringing the mistake to our attention.
If b and c are real and z complex ( i.e.: z=x+i y , with x and y real), the condition of convergence is y<0 as already said.
Of course, if all b, c, z are complex a more general condition has to be derived from :
[Real part of i((b/c)-z)] < 0
 

FAQ: Solve Exponential Integral - Get Help Now

What is an exponential integral?

An exponential integral is a mathematical function that involves the integration of an exponential function. It is often used in various fields of science and engineering to solve differential equations and other problems.

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