An Integral

[Erasmus00]

I know (by table) that the integral

 

[math] \int_0^{\infty} dx \frac{e^{(-ax)}-e^{(-bx)}}{x} = \ln(\frac{b}{a}) [/math]

 

Anyone know how you would go about doing such an integral? Is there some nice complex contour I'm missing?

 

For all the math wizzes out there, I could use a bit of help.

-Will

[sanctus]

Can't you apply the residual theorem separately to the two integrands in complex plane[math]e^{(-az)}z^{-1}\quad and \quad -e^{(-bz)}z^{-1}[/math], posing

Integrating on a countour from -R to -r (measure beingdx)then around the pole in zero (integrate from 0 to pi [math]re^{it}dt[/math] and the measure being i times the precedent parametrization of the countour), then from r to R (again dx) ant then from pi to 0 to go back from R to -R. then you show what happens in the limits R to infinity (what usually gives zero) and r to zero. Which may give you the answer but I'm not sure I haven't really tried.

[Erasmus00]

Can't you apply the residual theorem separately to the two integrands in complex plane[math]e^{(-az)}z^{-1}\quad and \quad -e^{(-bz)}z^{-1}[/math], posing

Integrating on a countour from -R to -r (measure beingdx)then around the pole in zero (integrate from 0 to pi [math]re^{it}dt[/math] and the measure being i times the precedent parametrization of the countour), then from r to R (again dx) ant then from pi to 0 to go back from R to -R. then you show what happens in the limits R to infinity (what usually gives zero) and r to zero. Which may give you the answer but I'm not sure I haven't really tried.

 

The trouble is that a. the residue at the pole is 1, which clearly isn't going to give me much in terms of a log, and the integration bound is from 0 to infinity, which means that I probably can't use a contour that stretches very far into the negative half of the real line.

-Will

[sanctus]

Yes, I had seen that residue was 1 and thought that it was a mistake in quick calculation...

 

I didn't think about the bounds sorry.

I have no other idea at the moment.

[Qfwfq]

No contour!!!!!! It isn't necessary to use residues and I'm a bit sceptical as to the method being useful to this case.

 

Now it's quite obvious that for a = b the integrand is identically zero, therefore so is the integral. It can be written as the difference between two values of a same function, but if one defines:

 

[math]f(a) = \int_0^{\infty}dx\,\frac{e^{-ax}}{x} [/math]

 

this isn't finite for any a, only differences are finite, so it's better to define (b = 0):

 

[math]f(a) = \int_0^{\infty}dx\,\frac{e^{-ax}-1}{x}[/math]

 

The trick is that, knowing f(0) = 0 and its derivative, f is determined by analyticity. It isn't difficult to find that

 

[math]\frac{d}{da}\int_0^{\infty}dx\,\frac{e^{-ax}-1}{x}=-\frac{1}{a}[/math]

 

which confirms the table's result, by starting with the definition of derivative and using de l'Hôpital's rule on the da --> 0 limit; at that point the integration in dx becomes trivial.

 

Notice that, strictly, [math]\ln\frac{b}{a}[/math] isn't the solution for all a and b. For a = b = 0 it isn't defined unless it's extended by continuity. For a and b of opposite signs neither it nor the integral are defined, but if both are negative the logarithm is and the integral isn't unless it is tweaked in an obvious way. Freaky, man, freaky! :D

[sanctus]

Wow, and you found it just like that?

But for the part where you said "it isn't difficult to find that" I have to look at when I'm not falling asleep...

[Erasmus00]

Thanks. Much appreciated.

-Will