# Improper integral.

I have example which asks to determine whether the improper integral converges or diverges, and if it converges I have to solve it.

I couldn't solve this, but I found a very descriptive solution, but there are parts that I don't understand, anyway, here's the integral:

$I=\int_0^{\frac{\pi}{2}} \ln(\sin x)dx$

Firstly, we have to determine it's convergence , and in the solution I have, it first uses integration by parts the following way:

$u=\ln(\sin x) \implies du=\cot x dx, dv=dx \Rightarrow v=x$

So now we have the integral:

$I=x\ln\sin x|_0^{\pi /2} -\int_0^{\pi/2} \frac{x}{\tan x}dx$

and here goes the part that I don't understand, it says now let's look at the integral we got after integration by parts, and if it converges it means that the integral we started with is convergent too. My question is: Why is that so? I mean, can we ignore the $x\ln\sin x|_0^{\pi /2}$ just like that? What actually happens with this part?

• Have you tried to evaluate the $x\ln(\sin(x))$ part? You'll need a limit for the $0$. Commented Apr 1, 2016 at 10:17
• This question has been asked several times on this website! One technique is to make a change of variables $u=\ sin x$ and then you can relate the new integral to beta function! Commented Apr 1, 2016 at 10:18
• $ln(0)$ is undefined Commented Apr 1, 2016 at 10:19

What you're reading is skipping over steps. Basically, in this example, we can ignore the $\left.x\ln(\sin(x))\right|_0^{\pi/2}$ because it is a constant (so the only part that can diverge is the integral.
To check that $\left.x\ln(\sin(x))\right|_0^{\pi/2}$ is constant, we can substitute $x=\frac{\pi}{2}$ to get $\frac{\pi}{2}\cdot\ln(\sin(\frac{\pi}{2}))=\frac{\pi}{2}\ln(1)=0$.
On the other hand, at $x=0$, the value is an indeterminate form ($0\cdot\infty$). We, however, can use l'Hopital's rule to solve this: \begin{align*} \lim_{x\rightarrow 0^+}x\ln(\sin(x))&=\lim_{x\rightarrow 0^+}\frac{\ln(\sin(x))}{\frac{1}{x}}\\ &=\lim_{x\rightarrow 0^+}\frac{\frac{\cos(x)}{\sin(x)}}{-\frac{1}{x^2}}\\ &=-\lim_{x\rightarrow 0^+}\frac{x^2\cos(x)}{\sin(x)}\\ &=-\lim_{x\rightarrow 0^+}x\cos(x)\cdot\frac{x}{\sin(x)}. \end{align*} The $\frac{x}{\sin(x)}$ part is an old friend and converges to $1$. The $\cos(x)$ converges to $1$, and the $x$ converges to $0$. Therefore, this limit converges to $0$.
Therefore, we can ignore the $x\ln(\sin(x))$ part because it happens to be a constant and not infinite (in particular, it is $0$).