For a class of Physics I need to compute the following integral:
$$\int_{-L}^{L}\mathrm{d}q\dfrac{\theta(\epsilon-bq)}{\sqrt{(\epsilon-bq)}}$$
and I truly have no idea on how to proceed. Note $\theta(\cdot)$ is Heaviside step function. Also, is there any systematic approach to solving this kind of integrals involving the $\theta$ function?
Best Answer
I will assume that all quantities $L,b,\epsilon$ are $>0$.
Moreover, as we have
$$\color{red}{0 <} \ \epsilon-bL \leq \epsilon-bq \leq \epsilon+bL \tag{1}$$
The inequality in red being compulsory for the square root to be defined.
Make the change of variables $r=\epsilon-bq$ giving $dr=-b dq \iff \mathrm{d}q=-\frac{1}{b}\mathrm{d}r$, then
$$\int_{q=-L}^{q=L}\mathrm{d}q\dfrac{\theta(\epsilon-bq)}{\sqrt{(\epsilon-bq)}}=\int_{r=\epsilon+bL}^{r=\epsilon-bL}-\frac{1}{b}\mathrm{d}r\dfrac{\theta(r)}{\sqrt{r}}=\frac{1}{b}\int_{r=\epsilon-bL}^{r=\epsilon+bL}\mathrm{d}r\dfrac{\theta(r)}{\sqrt{r}}=\frac{1}{b}\int_{r=\epsilon-bL}^{r=\epsilon+bL}\mathrm{d}r\dfrac{1}{\sqrt{r}}=$$
(we keep the same bounds due to (1)).
$$=\frac{1}{b}[2\sqrt{r}]_{r=\epsilon-bL}^{r=\epsilon+bL}$$
Now, it is easy to find the result.