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Math Help - [SOLVED] Integral of ln(sin(x))dx

  1. #1
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    [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.

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by Tofi007 View Post
    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.

    Thanks in advance!
    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.
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  3. #3
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    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
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  4. #4
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    Quote Originally Posted by Tofi007 View Post
    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 [tex]\ln[/Math] 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.
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  5. #5
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    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?.
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  6. #6
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    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
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  7. #7
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    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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