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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?

sai-kartik
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Elementarium
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2 Answers2

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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.

Jean Marie
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  • I am looking for a solution involving the $\theta$ function. Also, I do not see why you remove the $\theta$ in the last step. – Elementarium Apr 29 '20 at 14:32
  • I have remove the $\theta$ in the last step because the region of integration is included into the positive reals. But you can keep it if you want... – Jean Marie Apr 29 '20 at 14:38
  • I have added the (almost final) result which complies with your result (OK by including $\theta$). – Jean Marie Apr 29 '20 at 14:40
  • Yes, I understand your procedure, but I have not imposed any conditions on the given values. Actually, wcan take $b$ and $L$ positive, but about $\epsilon$ I have no restrictions – Elementarium Apr 29 '20 at 14:42
  • Also, is what I stated on a comment to the other answer about the primitive of the function correct? – Elementarium Apr 29 '20 at 14:44
  • Yes, this primitive function is correct, just as the answer without $\theta(x)$ is correct if we are in the $>0$ domain. Besides, about the sign of $\epsilon$, as I don't know what $\epsilon$ represents "physically", I cannot but assume that taking it positive hopefully will not harm the whole process... – Jean Marie Apr 29 '20 at 17:08
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Since Heaviside function is a piecewise function: $$ \theta(x)=\begin{cases}1&x\ge0,\\0&x<0.\end{cases} $$ and composition of functions involving piecewise functions is also a piecewise function, you can rewrite you integral as (assuming $b>0$): $$ \int_{-L}^{L}dq\begin{cases}0&\epsilon-bq <0,\\\frac1{\sqrt{\epsilon-bq}}&\epsilon-bq \ge 0,\end{cases} = \int_{-L}^{L}dq\begin{cases}0&q > \epsilon/b,\\\frac1{\sqrt{\epsilon-bq}}&q \le\epsilon/b.\end{cases} $$ Finally integral of piecewise function can be seen as a sum of integrals of each piece. If the integral is definite, you also want to keep in mind that the transition of piecewise function may happen outside the bounds of integration: $$ \begin{cases} \int_{-L}^{L} \frac{dq}{\sqrt{\epsilon-bq}} &L \le\epsilon/b, \\ \int_{-L}^{\epsilon/b} \frac{dq}{\sqrt{\epsilon-bq}}+\int_{\epsilon/b}^L 0dq &-L \le\epsilon/b<L,\\ \int_{-L}^L 0\,dq & \epsilon/b<-L \end{cases} $$

Vasily Mitch
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  • This is what I was looking for, yet the answer in my notes is $\frac{2}{\beta}(\sqrt{\epsilon+bL}\theta(\epsilon+bL)-\sqrt{\epsilon-bL}\theta(\epsilon-bL))$. I have been doing calculations and cannot figure out how to arrive to this solution. Wolfram alpha states that the primitive of $x^{-1/2}\theta(x)$ is 2\sqrt(x)\theta(x)+C$ and this would resolve the problem, but I am still not convinced nor do I know how that primitive is computed – Elementarium Apr 29 '20 at 14:29
  • If $\epsilon - bq <0$ the integral is undefined : we shouldn't consider this case. Moreover you should say that you make the implicit assumption that $b>0$ when you consider as equivalent $\epsilon - bq <0$ and $q>\epsilon/b$. – Jean Marie Apr 29 '20 at 14:30