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Math Help - Definitely Integral

  1. #1
    Lord of certain Rings
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    Wink Definitely Integral

    To all Integral lovers,

    In the Spirit of Integration....here is a popular integral

    \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan x} \right)dx}

    Well Enjoy,
    Isomorphism

    P.S: Elegant methods score can score a lot of 'thank you's
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    Lord of certain Rings
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    Quote Originally Posted by ThePerfectHacker View Post
    I will wait for some different way of doing it. I have a solution that looks more elegant... but it is almost the same as you did....
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    Lord of certain Rings
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    OK i will post my solution since nobody else replied

    I = \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan x} \right)dx}

    I = \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan \left(\frac{\pi}4 - x\right)} \right)dx}

    I =\int\limits_0^{\frac {\pi }{4}} \ln \frac2{1 + \tan x} dx

    I = \int\limits_0^{\frac {\pi }{4}}\ln 2 \,dx - I

    So I = \frac{\pi}8 \ln2


    Part 2:
    \int^1_0\frac{x^4(1-x)^4}{1+x^2} dx

    This is a funny integral... not hard... but the answer is interesting
    People who solve it can comment on the answer rather than showing the answer
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
    OK i will post my solution since nobody else replied

    I = \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan x} \right)dx}

    I = \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan \left(\frac{\pi}4 - x\right)} \right)dx}

    I =\int\limits_0^{\frac {\pi }{4}} \ln \frac2{1 + \tan x} dx

    I = \int\limits_0^{\frac {\pi }{4}}\ln 2 \,dx - I

    So I = \frac{\pi}8 \ln2


    Part 2:
    \int^1_0\frac{x^4(1-x)^4}{1+x^2} dx

    This is a funny integral... not hard... but the answer is interesting
    People who solve it can comment on the answer rather than showing the answer
    Haha...but how close is that approximation to the real thing...is it less or more making this positive or negative haha
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    Lord of certain Rings
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    Quote Originally Posted by Mathstud28 View Post
    Haha...but how close is that approximation to the real thing...is it less or more making this positive or negative haha
    Hmm actually this is used to approximate 'it' by the use of sandwiching the integral between two series :P
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    Lord of certain Rings
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    I thought of converting this thread into an "integral collection of my life". So I will be posting integrals here that seem nice to me. Anyone who thinks (s)he has a nice solution, can of course post them here... Thank you

    The next one in the set:

    Part 3:
    \int_0^\infty \frac{dx}{1+x^4}
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    Quote Originally Posted by Isomorphism View Post
    Part 3:
    \int_0^\infty \frac{dx}{1+x^4}
    If I remember correctly: \int_0^{\infty} \frac{dx}{x^n+1} = \frac{(\pi/n)}{\sin (\pi/n)} where n\geq 2.
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    Lord of certain Rings
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    Quote Originally Posted by ThePerfectHacker View Post
    If I remember correctly: \int_0^{\infty} \frac{dx}{x^n+1} = \frac{(\pi/n)}{\sin (\pi/n)} where n\geq 2.
    Can you prove that without complex analysis(that is without using residues)?
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
    Can you prove that without complex analysis(that is without using residues)?
    Do you mean by decomposition on the bottom or do you mean by using something else?
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  11. #11
    Lord of certain Rings
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    \int_0^\infty \frac{dx}{1+x^4} can be solved without complex numbers. I used only pure substitution and algebraic tricks

    P.S: Unfortunately what I did only holds for this problem . It does not work for TPH's generalization.
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    Quote Originally Posted by Isomorphism View Post

    Can you prove that without complex analysis(that is without using residues)?
    Combine Gamma & Beta functions plus the Euler Reflection Formula.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Combine Gamma & Beta functions plus the Euler Reflection Formula.
    huh?
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    Let \int_{0}^{\infty }{\frac{1}{1+x^{n}}\,dx}. Make the substitution z=x^n, hence

    \begin{aligned}<br />
\int_{0}^{\infty }{\frac{1}{1+x^{n}}\,dx}&=\frac{1}{n}\int_{0}^{\in  fty }{\frac{z^{1/n-1}}{1+z}\,dz}\\ <br />
 &=\frac{1}{n}\beta \left(\frac{1}{n},1-\frac{1}{n}\right)\\ <br />
 &=\frac{1}{n}\cdot \frac{\Gamma \left( n^{-1}\right)\Gamma\left( 1-n^{-1} \right)}{\Gamma (1)}\\ <br />
&=\frac{1}{n}\cdot \frac{\pi}{\sin\dfrac{\pi}{n}}=\frac{\pi }{n\sin\dfrac{\pi }{n}}.\quad\blacksquare<br />
\end{aligned}

    Note aside: I've heard that the Reflection Formula can be only proved via complex analysis. It's a dream to see a solution with elementary tools.
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  15. #15
    Lord of certain Rings
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    Here is the latest update of "Definitely Integral"


    Compute I:

    \boxed{I = \int_0^{\pi/4}\ln(\sin x) \cdot \ln(\cos x)\,dx}






    P.S:I think its quite hard
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