Wolfram Alpha isn't able to calculate this integral (I don't have mathematica, but I have Wolfram Pro).
$$\int_{0}^{a} \frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx \ \ \ , \ b>a$$
This is for a physics problem. I'd appreciate either a solution or the knowledge that the integral is non-soluble (which would indicate that I need to find some symmetry that I haven't seen yet). Thanks!
Best Answer
I was able to specify the closed-form with Maple.
$$\int_{0}^{a} \frac{1}{\sqrt{(x-a)^2+(x-b)^2}} \ dx$$ equals to $$\frac {\sqrt 2}2 \left(\ln \left( \left( \sqrt {2}+2\,{\operatorname{csgn}} \left( a-b \right) \right) \left( a-b \right) \right) - \ln \left(2\,\sqrt {{a}^{2}+{b}^{2}}-\sqrt {2}\left(a+b\right) \right) \right), $$
where $\operatorname{csgn}$ is the complex signum function.
If we assume that $b>a$, then we could simplify it into the form
$$\frac {\sqrt 2}2 \left( \ln \left( b-a \right)+\ln \left( \sqrt {2}-1 \right) -\ln \left( \sqrt {2} \sqrt {{a}^{2}+{b}^{2}}-a-b \right) \right).$$