I'm having a tough time with this integral (Magnetic Vector Potential)

Question

Wolfram Alpha isn't able to calculate this integral (I don't have mathematica, but I have Wolfram Pro).

$\int_{0}^{a} 1/\sqrt{(x-a)^2+(x-b)^2}dx$ $;b>a$

This problem comes from solving for the Magnetic Vector Potential caused by a current $I$ running in a triangular loop from the origin to $\textbf{c}=(a,0,0)$ to $\textbf{d}=(0,a,0)$ then back to the origin at the point $(x,0,0)$. Assume x>a. The integral given above describes the contribution from the $\bf{c}$ to $\bf{d}$ segment.

This might help: $(x-a)^2 + (x-b)^2 = 2[x-(a+b)/2]^2 +(a^2 + b^2)/2 - ab$ — Pulsar, Nov 10 '14 at 20:29
Thanks for the reminder about the relationship between b and a. — Zane Dufour, Nov 10 '14 at 20:31
Assuming $a>0$, the integral is $\frac{1}{\sqrt{2}} \left(-\mathrm{arcsinh}(1)+\log \left(\frac{a-b}{a+b-\sqrt{2(a^2+b^2)}}\right) \right)$. — JPhy, Nov 10 '14 at 20:32
Possible duplicate by OP: http://math.stackexchange.com/q/1015685/11127 — Qmechanic, Nov 10 '14 at 21:55

score 0 · Accepted Answer · answered Nov 10 '14 at 20:33

I suppose this could work. Hope i have not made any arithmetic error

Complete squares in the denominator. Please cross check though

$\sqrt{2 \left(\left(x-\frac{a+b}{2}\right)^2 + \left(\frac{a-b}{2}\right)^2\right)}$

Now substitute $t = x - \frac{a+b}{2}$ and it reduces to integral of form $\frac{dt}{t^2+r^2}$, where $r = \frac{a-b}{2}$. This is a standard integral which reduces to logarithm or $sinh^-1$ depending upon sign of $a-b$

I'm having a tough time with this integral (Magnetic Vector Potential)

1 Answers1