Solution of an integral containing vectors

Question

I'm currently trying to solve the integral: $$ I(\vec{a},\vec{b})=4\pi\int\limits_0^1\frac{\mathrm{d}u}{1-(\vec{a}u+\vec{b}(1-u))^2}, $$ but I can't seem to find a good starting point. I know that if $\vec{a}$ and $\vec{b}$ are scalars this integral would be easilly solved by a partial fraction decomposition (where the relation $x^2-y^2=(x-y)(x+y)$ is really helpful), yielding an answer in terms of the natural logarithm.

Right now however I am stuck with a vector relation, which doesn't make the calculations any easier. I tried the partial fraction decomposition, this doesn't make things easier. I also tried writing out the denominator, but this also isn't that helpful.

I was wondering if there were general ways to tackle these kind of problems and/or if there are any tips for this integral ?

The vectors $\vec{a}$ and $\vec{b}$ are 3D, this can for example be seen by the fact that the factor $4\pi$ follows from the 3D solid angle $$ 4\pi=\int\limits_0^{2\pi}\mathrm{d}\varphi\int\limits_{0}^\pi\sin(\theta)\mathrm{d}\theta. $$

Answer 1

As you said, integrals are best tackled with scalars. Thus the most straightforward way to solve these types of problems (unless there's some other nice geometric property about the vectors to make it easier) is to deal with the individual components. Instead of $\vec a$ and $\vec b$, deal with $a_x, a_y, b_x, b_y$. Then we have $$(\vec a u + \vec b(1-u))^2 = (a_x u+b_x(1-u),a_y u+b_y(1-u))^2 = b_x^2 + b_y^2 + 2(a_x b_x - b_x^2 + a_y b_y - b_y^2) u + (a_x^2 + a_y^2 - 2 a_x b_x + b_x^2 - 2 a_y b_y + b_y^2) u^2$$ Which is -- from the perspective of $u$, the variable we're integrating -- just a quadratic. So the integral becomes $$\int \frac{du}{(1-b_x^2+b_y^2) + 2(b_x^2 + b_y^2 - a_x b_x - a_y b_y) u + (2 a_x b_x + 2 a_y b_y - b_x^2 - b_y^2 - a_x^2 - a_y^2 ) u^2}$$ or perhaps more succinctly (and geometrically) $$\int \frac{du}{(1-|b|^2) + 2(|b|^2-a\cdot b)u + (2a\cdot b-|b|^2-|a|^2)u^2}$$ Then you can apply that $$\int_0^1 \frac{du}{x + y u + z u^2} = \frac{2\left(\arctan\left(\frac{y+2z}{\sqrt{4xz-y^2}}\right)-\arctan\left(\frac{y}{\sqrt{4xz-y^2}}\right)\right)}{\sqrt{4xz-y^2}}$$ and plug in those values for $x, y, z$ to get your integral.

Answer 2

First write your norm $F(u) = \left\Vert u \vec{a} u + (1-u)\vec{b}\right\Vert^2$ in the form of a quartic in $u$, as $A u^2 + B u + C$ (Use the expression of vectors $\vec{a}$, $\vec{b}$ in orthogonal coordinates if needed). Then it should be straightforward (there is a general, well-known method for integrating $1/\mathrm{quartic}$).