3
$\begingroup$

I'm solving a physics problem, and at some point I need to solve this integral

$$\iint\limits_S \frac{1}{x+2a} dxdy$$

where

$$S=\{(x,y,z):x^2+y^2\leq a^2, \ x\geq0, \ z=0\}$$

So $S$ is the right semicircle corresponding to the circle centered at the origin with radius $a$. This seemed a bit difficult so I decided to calculate the integral over the quarter circle in the first quadrant (let's call this surface $S'$) and then multiply that by two to get the value of the whole surface integral. Doing this, I got

$$2\iint\limits_{S'} \frac{1}{x+2a} dxdy=2\int\limits_0^a dy\int\limits_0^{\sqrt{a^2-y^2}}\frac{1}{2a+x}dx=2\int\limits_0^a\ln\left(\sqrt{a^2-y^2}+2a\right)-\ln(2a)\ dy=\int\limits_0^a 2\ln\left(\sqrt{a^2-y^2}+2a\right)\ dy \ - a\ln(2a)$$

And this function doesn't seem to have a primitive, or at least it is very hard to find it (I looked up in some tables and found nothing). The problem shouldn't be so hard so I expect that this integral can be calculated in a much easier way. Any ideas?

$\endgroup$
  • $\begingroup$ this kind of integrals was done very often on this site...look around carefully and i am sure you will find something helpful $\endgroup$ – tired Oct 10 '16 at 13:46
  • $\begingroup$ You have indicated an infinite 3-dimensional region, not a surface. Please double-check. $\endgroup$ – Ted Shifrin Oct 10 '16 at 14:05
  • $\begingroup$ @TedShifrin Edited. Sorry about that $\endgroup$ – Tendero Oct 10 '16 at 14:09
  • 3
    $\begingroup$ Integrating $y$ first and using the trig substitution $x = a\sin\theta$ leads to$$\int_{0}^{a} \frac{2\sqrt{a^{2} - x^{2}}\, dx}{2a + x} = \int_{0}^{\pi/2}\frac{2a\cos^{2}\theta\, d\theta}{2 + \sin\theta},$$which at least has an elementary primitive (though it may be messy; haven't pursued this further). $\endgroup$ – Andrew D. Hwang Oct 10 '16 at 14:42
  • $\begingroup$ @AndrewD.Hwang nice idea! the result becomes quite elementary wolframalpha.com/input/… $\endgroup$ – tired Oct 10 '16 at 14:56
1
$\begingroup$

Integrating $y$ first and using the trig substitution $x = a\sin\theta$ leads to $$ \int_{0}^{a} \frac{2\sqrt{a^{2} - x^{2}}\, dx}{2a + x} = \int_{0}^{\pi/2} \frac{2a\cos^{2}\theta\, d\theta}{2 + \sin\theta}. \tag{1} $$ The tangent half-angle substitution $t = \tan\frac{\theta}{2}$ gives $$ \cos\theta = \frac{1 - t^{2}}{1 + t^{2}},\qquad \sin\theta = \frac{2t}{1 + t^{2}},\qquad d\theta = \frac{2\, dt}{1 + t^{2}}, $$ upon which (1) becomes $$ 2a\int_{0}^{1} \frac{\left(\dfrac{1 - t^{2}}{1 + t^{2}}\right)^{2}}{2 + \dfrac{2t}{1 + t^{2}}} \cdot \frac{2\, dt}{1 + t^{2}} = 2a\int \frac{(1 - t^{2})^{2}}{(1 + t^{2})^{2}(1 + t + t^{2})}\, dt. \tag{2} $$ Partial fractions is a bit laborious (six equations, six unknowns); for the record, the decomposition is $$ \frac{(1 - t^{2})^{2}}{(1 + t^{2})^{2}(1 + t + t^{2})} = \frac{A_{1}t + B_{1}}{(t^{2} + 1)^{2}} + \frac{A_{2}t + B_{2}}{t^{2} + 1} + \frac{A_{3}t + B_{3}}{t^{2} + t + 1}, $$ and the augmented matrix of the resulting linear system for $(A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3})$ is $$ \left[\begin{array}{@{}rrrrrr|r@{}} 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 2 & 1 & 2 & 0 & 0 \\ 1 & 1 & 1 & 2 & 0 & 2 & -2 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 \\ \end{array}\right]. $$ The end result is easily checked to be $$ \frac{(1 - t^{2})^{2}}{(1 + t^{2})^{2}(1 + t + t^{2})} = -\frac{4t}{(t^{2} + 1)^{2}} + \frac{4}{t^{2} + 1} - \frac{3}{(t + \frac{1}{2})^{2} + \frac{3}{4}}. $$ Consequently, (2) becomes $$ 2a\left[\frac{2}{t^{2} + 1} + 4\arctan t - 2\sqrt{3} \arctan\bigl(\tfrac{2}{\sqrt{3}}(t + \tfrac{1}{2})\bigr)\right]\bigg|_{0}^{1} = 2a\left[-1 + \pi + \frac{\pi}{\sqrt{3}}\right]. $$

$\endgroup$
  • $\begingroup$ The fact that Wolfram Alpha spits out an answer in a few seconds tends to make one feel like Kyûzô in The Seven Samurai.... $\endgroup$ – Andrew D. Hwang Oct 10 '16 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.