Solve Exponential Integral - Get Help Now

[Skullmonkee]

Hi i was wondering if anyone could help me with a fairly simple integration.

\int exp(iab/c)exp(-iaz) da

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.

 

[Pengwuino]

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, e^{ix} = cos(x) + isin(x) and you'll end up trying to take a limit at infinity which does not exist for cosines and sines.

 

[ACPower]

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 <br /> \int cos(aK) da + i\int sin(aK) da = 0<br />
since integrating a sine or cosine gives zero, since there are equal positive and negative areas for every cycle.

 

[Pengwuino]

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.

 

[JJacquelin]

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))

 

[adriank]

The correct answer follows:

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

Pengwuino and ACPower assume that b/c - z is real, but if that is not the case, the integrand may not be purely sinusoidal (that is, k above may have nonzero real part). JJacquelin gives an incorrect value for the integral (there is a missing c).

 

[JJacquelin]

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