A Frusterating Double-Integral

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SUMMARY

The discussion centers on solving the double integral \[ \int \int \frac{1}{[(x_1-x_2)^2+a^2 ]^{1/2}} dx_1 dx_2 \] with bounds of [-L/2, L/2] for both integrals. The participant expresses difficulty in finding a suitable approach, considering trigonometric substitutions and the potential need for integral tables. A suggested substitution is \[ x_1-x_2=a f(u) \] which simplifies the integral to a form involving \[ \frac{1}{\sqrt{f(u)^2+1}}. \

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Homework Statement



Integrate:

\int \int \frac{1}{[(x_1-x_2)^2+a^2 ]^1^/^2} dx_1 dx_2

(The bounds on the first integral are [-L/2, L/2] and the second integral's bounds are the same.)

Homework Equations



N/A

The Attempt at a Solution



I stared at this problem long and hard, and was hoping to use some sort of a trigonometric substitution, but nothing came to mind. I have a suspicion that I might need integral tables to solve this problem, but I might be jumping the gun on this.

I feel like the 'a' variable is throwing a real wrench in the system. If anyone could maybe nudge me in the right direction (if this is solvable without consulting a table) I'd be very appreciative.
 
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To get rid of the a we could use a substitution x1-x2=a f(u). This would allow you to move the a in front of the integral sign. We then have a relation of the form 1/sqrt(f(u)^2+1). Now you can find a trigonometric or hyperbolic function for f(u) that obeys f(u)^2+1=g(u)^2. Do you know which one?
 

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