Originally Posted by

**Furyan** Hi ILikeSerena, I certainly feel very close to a black hole, but I'm afraid I can't resist the field strength

. I figure since I've got the integral it should just be plain sailing. What do you mean by singularities?

Thank you

Actually, now that I'm taking a better look, it's even worse. Your integral does not exist.

Your given answer is wrong.

The problem is that $\displaystyle \cot 7x$ is not defined for all the points in the range.

It has so called singularities in $\displaystyle x=0, \frac \pi 7, \frac {2\pi} 7, \frac {3\pi} 7$.

It means that you have to split your integral up at these singularities into so called improper integrals (see included wiki link).

Your integral expression $\displaystyle 7 \cot^2(7x)$ has infinitely high mountains in these singularities.

Now it could still be possible that these mountains have a finite area, but in this case they don't.

If we split up your integral at the singularities, we get:

$\displaystyle \int_{\pi / 4}^{\pi / 2} 7 \cot^2(7x) dx = \int_{\pi / 4}^{2\pi / 7} 7 \cot^2(7x) dx + \int_{2\pi / 7}^{3\pi / 7} 7 \cot^2(7x) dx + \int_{3\pi / 7}^{\pi / 2} 7 \cot^2(7x) dx$

Let's pick the first term:

$\displaystyle \int_{\pi / 4}^{2\pi / 7} 7 \cot^2(7x) dx$$\displaystyle = \lim_{b~ \uparrow~ {2\pi / 7}} \int_{\pi / 4}^{b} 7 \cot^2(7x) dx $

$\displaystyle = \lim_{b~ \uparrow~ {2\pi / 7}} (-\cot(7x)-7x)|_{\pi / 4}^b$

$\displaystyle =\text{does not exist}$

This improper integral does not exist, because $\displaystyle \cot(7\cdot {2\pi \over 7}) = \cot(2 \pi)$ is not defined.

None of these improper integrals exist, so your integral does not exist either $\displaystyle \Box$.