I stumbled upon this integral while doing Solid-state physics homework:
$$\int_\mu^{2t} \frac{1}{E+2\mu -2\epsilon}\frac{1}{\sqrt{4t^2-\epsilon^2}} \mathrm d\epsilon$$
where $E<0$, $\mu \in (-2t,2t)$ and $t>0$.
I tried plugging it into Mathematica:
Integrate[1/(e + 2 \[Mu] - 2 x) 1/Sqrt[4 t^2 - x^2], {x, \[Mu], 2 t},
Assumptions -> -2 t < \[Mu] < 2 t && t > 0 && e < 0]
and got the following answer:
$$-\frac{1}{2} \pi \sqrt{\frac{1}{E^2+4 E \mu +4 \mu ^2-16 t^2}}-\frac{\arctan\left(\frac{\mu (E+2 \mu )-8 t^2}{\sqrt{\left(4 t^2-\mu ^2\right) \left(E^2+4 E \mu +4 \left(\mu ^2-4 t^2\right)\right)}}\right)}{\sqrt{E^2+4 E \mu +4 \left(\mu ^2-4 t^2\right)}}$$
This result gives correct "physical" interpretation, so now I would like to know how this integral was
calculated.
My first guess was to use substitution $\epsilon = 2t\sin{x}$ and transform this integral to
$$\int_{\arcsin{\frac{\mu}{2t}}}^{\frac{\pi}{2}} \frac{\mathrm dx}{E+2\mu – 4t\sin{x}}
$$
Here I thought about trying to evaluate indefinite integral by doing the standard substitution $u = \tan{\frac{x}{2}}$, however I hesitated after reading this comment.
Is there any other way to obtain the result in closed form like the one obtained from Mathematica?
Best Answer
If you just ask Mathematica to compute the antiderivative without any assumption, you should receive $$I=\int \frac{dx}{\sqrt{4 t^2-x^2} (e+2 \mu -2 x)}=$$ $$\sqrt{-e^2-4 e \mu -4 \mu ^2+16 t^2}\,I=-\log (e+2 \mu -2 x)+$$ $$\log \left(\sqrt{4 t^2-x^2} \sqrt{-e^2-4 e \mu -4 \mu ^2+16 t^2}-e x+8 t^2-2 \mu x\right)$$ and I suppose that the arctangent comes from the combination of the logarithms.