# Thread: [SOLVED] Integral of ln(sin(x))dx

1. ## [SOLVED] Integral of ln(sin(x))dx

Hi I'm kinda new here, but you may be my only hope, I've got 4 extremely hard problems. Here's one of them:

$\int_0^{\pi}\ln(\sin x) \, \mathrm{d}x$

I've been trying to solve these for more than 8 hours now, but nothing seems to get me anywhere.

2. Originally Posted by Tofi007
Hi I'm kinda new here, but you may be my only hope, I've got 4 extremely hard problems. Here's one of them:

$\int_0^{\pi}\ln(\sin x) \, \mathrm{d}x$

I've been trying to solve these for more than 8 hours now, but nothing seems to get me anywhere.

There's a trick for this one: split the integral into $\int_0^{\pi/2}(\cdots)+\int_{\pi/2}^\pi(\cdots)$ and, in the second integral, procede to the change of variable $y=\frac{\pi}{2}-x$. As you will see, things will simplify.

3. Thanks a lot! Now that's a trick!
I don't want to be shameless, but I'm really hopeless with these problems.

These 2 are related in some way I think:

1. Prove that this is, or isn't convergent:

$\int_0^{\infty}x^2\cos (e^x)\, \mathrm{d}x$

2. For what p, and q is this convergent, and for what p, and q is this absolute convergent?

$\int_0^{\infty}x^p\sin (x^q)\, \mathrm{d}x$

4. Originally Posted by Tofi007
Thanks a lot! Now that's a trick!

1. Prove that this is, or isn't convergent:
Wait, what you needed is to prove that the integral is convergent? Then my trick is not the good one. Well, you still can split on $[0,\frac{\pi}{2}]$ and $[\frac{\pi}{2},\pi]$, but then this is the substitution $y=\pi-x$ that would be nice, since you end up with $2\int_0^{\pi/2} \ln\sin x\,dx$, and you only have to check that the function $\ln\sin$ is integrable on $[0,\pi/2]$.
To do this, notice that this function is continuous on $(0,\frac{\pi}{2}]$, and $|\ln\sin x|\sim |\ln x|$ as $x\to 0$, because $\sin x\sim x$ converges to 0 (and using a property of $$\ln$$ and asymptotic equivalence). Then, since $\int_0^1 \ln x\,dx$ converges, $\int_0^{\pi/2} \ln\sin x\,dx$ converges as well.

The trick I gave relates to the computation of the integral (once you know it exists). Using it, you get $\int_0^\pi \ln\sin x\,dx=\int_0^{\pi/2}\ln (\sin x\cos x)\,dx=-\frac{\pi}{2}\ln 2 + \int_0^{\pi/2} \ln \sin{2x}\,dx$, and the substitution $z=2x$ gives $\int_0^\pi \ln\sin x\,dx=-\frac{\pi}{2}\ln 2 + \frac{1}{2}\int_0^{\pi} \ln \sin x\,dx$, and you deduce the value for your integral. But this assumes you proved its existence first.

5. If I may put in my 2 cents. It ain't much.

$I=\int_{0}^{\pi}ln(sin(x))dx$

Let $sin(x)=2sin(\frac{x}{2})cos(\frac{x}{2})$

Then we get:

$I=\int_{0}^{\pi}ln(2sin(\frac{x}{2})cos(\frac{x}{2 }))dx$

$I=\int_{0}^{\pi}ln(2)dx+\int_{0}^{\pi}ln(sin(\frac {x}{2}))dx+\int_{0}^{\pi}ln(cos(\frac{x}{2}))dx$

Now, use the sub $u=\frac{x}{2}$

$I={\pi}ln(2)+2\int_{0}^{\frac{\pi}{2}}ln(sin(u))du +2\int_{0}^{\frac{\pi}{2}}ln(cos(u))du$

Now, use the identity: $cos(u)=sin(\frac{\pi}{2}-u)$

Make the sub: $t=\frac{\pi}{2}-u$ in the integral with cos(u) and finish. Is that OK?.

6. I knew this had been done before. See here and go down to post #50:

http://www.mathhelpforum.com/math-he...tegrals-2.html

This link has many fun tough integrals

7. Here's one that's even tougher. I reckon it depends on how one looks at it whether it's tougher or not.

$\int_{0}^{\frac{\pi}{2}}x\cdot ln(sin(x))dx$

Actually, if we try hard, we can probably find a general form of:

$\int_{0}^{\frac{\pi}{2}}x^{n}\cdot ln(sin(x))dx$

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