I have a problem that has kept me awake for several nights XD

If I integrate this (where $K$ and $S$ are constants;$K$ is small and $S$ can take values between 0 and 1): $$\int_{0}^{\pi}\frac{1}{\sin^2t}\left[\left(12\cos^2t+K^2\sin^2t\right)^{\frac{S+3}{2}}-\left(12\cos^2t\right)^{\frac{S+3}{2}}\right]\mathrm{d}t$$ by parts, I get this: $$=\left[ \left(\left(12\cos^2t+K^2\sin^2t\right)^{\frac{S+3}{2}}-\left(12\cos^2t\right)^{\frac{S+3}{2}}\right)(-\cot t)\right]^\pi_0-\int_{0}^{\pi}[...]$$ My question is: does the term $$\left[ \left(\left(12\cos^2t+K^2\sin^2t\right)^{\frac{S+3}{2}}-\left(12\cos^2t\right)^{\frac{S+3}{2}}\right)(-\cot t)\right]^\pi_0$$ vanish when the limits of integration are applied? Someone$^*$ says yes, but why?

Thanks in advance!


$^*$ here is a statement about it from a French-speaking mathematician:

enter image description here

It reads (more or less)

One can integrate by parts $$\int\frac{dt}{\sin^2t}[\cdots]=$$ This expression simplifies.

  • Indeed, the terms in brackets in the first line cancel, due to $\sin^2t/\sin t$.

Edit: Can be used L'Hospital's rule to check that?

  • $\begingroup$ Consider the Laurent series for $\cot z$. I haven't worked it out, but it's the first thing I would do. $\endgroup$
    – saulspatz
    Commented Mar 5, 2021 at 3:38
  • $\begingroup$ Thanks, but the problem would be, using the Laurent series for cot(t), the lower limit: you'd get 1/0. $\endgroup$ Commented Mar 5, 2021 at 3:43
  • $\begingroup$ Those exponents of $\frac{S+3}2$ worry me. Even setting this fraction to some nice letter $m$, say, makes the whole much less messy-looking, but I am still worried. $\endgroup$
    – Lubin
    Commented Mar 5, 2021 at 5:50
  • $\begingroup$ Thanks. In fact, $S$ can take values between 0 and 1. $\endgroup$ Commented Mar 5, 2021 at 16:12

1 Answer 1


I've worked this problem out and found, it's the result of an improper integral, where L'Hospital's rule can be applied. That is, if $\mathrm{cot}\;t=\frac{\mathrm{cos}\;t}{\mathrm{sin}\;t}$, we'll get: $$\underset{t\to \pi }{\mathrm{lim}} \left\lbrack {\left(12\;{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+3}{2}} -{\left(12\;{\mathrm{cos}}^2 \;t\right)}^{\frac{s+3}{2}} \right\rbrack \left(-\frac{\mathrm{cos}\;t}{\mathrm{sin}\;t}\right)$$ $$=\underset{t\to \pi }{\mathrm{lim}} \frac{\left\lbrack {\left(12\;{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+3}{2}} -{\left(12\;{\mathrm{cos}}^2 \;t\right)}^{\frac{s+3}{2}} \right\rbrack \left(-\mathrm{cos}\;t\right)}{\mathrm{sin}\;t}$$ $$\overset{L\prime \mathrm{Hôpital}}{=} \underset{t\to \pi }{\mathrm{lim}} \frac{\left(\frac{s+3}{2}\right)\left\lbrack {\left(12{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+1}{2}} \left(K^2 -12\right)+12{\left(12{\mathrm{cos}}^2 \;t\right)}^{\frac{s+1}{2}} \right\rbrack \left(\mathrm{sin}\;2t\right)\left(-\mathrm{cos}\;t\right)+\cdots }{\mathrm{cos}\;t}$$ $$\cdots \frac{\left\lbrack {\left(12{\mathrm{cos}}^2 \;t+K^2 {\mathrm{sin}}^2 \;t\right)}^{\frac{s+3}{2}} -{\left(12{\mathrm{cos}}^2 \;t\right)}^{\frac{s+3}{2}} \right\rbrack \left(\mathrm{sin}\;t\right)}{\mathrm{cos}\;t}$$ Now, if $\mathrm{sin}\;2\pi =0$, we'll get: $$=\frac{\left(\frac{s+3}{2}\right)\left\lbrack {\cdots} \right\rbrack \left(0\right)\left(1\right)+\left\lbrack {\cdots} \right\rbrack \left(0\right)}{-1}=0$$ Similarly, if $\mathrm{sin}\;0 =0$, we'll get: $$=\frac{\left(\frac{s+3}{2}\right)\left\lbrack {\cdots} \right\rbrack \left(0\right)\left(-1\right)+\left\lbrack {\cdots} \right\rbrack \left(0\right)}{1}=0$$ :-)

P.S. In case that somebody needs it, I can explain how the derivative was made.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .