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Math Help - Definitely Integral

  1. #16
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    Quote Originally Posted by Isomorphism View Post
    \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.

    While I was playing with \int_0^\infty \frac{dx}{1+x^4}, I found out an interesting way to prove \int_0^\infty \frac{1 - x^2}{1+x^4}\, dx = 0

    With x \to \frac1{x} substitution:

    \int_0^\infty \frac{dx}{1+x^4} = \int_0^\infty \frac{x^2 dx}{1+x^4}

    Its wonderful. Isnt it?
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  2. #17
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    Here is the latest one:

    \boxed{I = \int_0^1 \int_0^x \frac{\ln (1 - y)}{y(y-1)}  \, dy \, dx}

    I invented it after seeing a recent problem
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  3. #18
    Eater of Worlds
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    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
    We can do this one by just dividing. It's not that hard but the solution is cool.

    \int\left[x^{6}-4x^{5}+5x^{4}-4x^{2}-\frac{4}{x^{2}+1}+4\right]dx

    Which gives us the solution of \frac{22}{7}-{\pi}
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  4. #19
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Isomorphism View Post
    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
    First note that: <br />
\int_0^{\tfrac{\pi }<br />
{2}} {\ln \left[ {\sin \left( \theta  \right)} \right] \cdot \ln \left[ {\cos \left( \theta  \right)} \right]d\theta }  = 2 \cdot \int_0^{\tfrac{\pi }<br />
{4}} {\ln \left[ {\sin \left( \theta  \right)} \right] \cdot \ln \left[ {\cos \left( \theta  \right)} \right]d\theta } <br />

    Recall the identity: {\rm B}\left( {x,y} \right) = 2 \cdot \int_0^{\tfrac{\pi }<br />
{2}} {\sin ^{2x - 1} \left( \theta  \right) \cdot \cos ^{2y - 1} \left( \theta  \right)d\theta } <br />

    Thus: <br />
\left. {\frac{{\partial ^2 {\rm B}\left( {x,y} \right)}}<br />
{{\partial x \cdot \partial y}}} \right|_{x = y = \tfrac{1}<br />
{2}}  = 8 \cdot \int_0^{\tfrac{\pi }<br />
{2}} {\ln \left[ {\sin \left( \theta  \right)} \right] \cdot \ln \left[ {\cos \left( \theta  \right)} \right]d\theta } <br />

    Which can be computed using the identity: <br />
{\rm B}\left( {x,y} \right) = \frac{{\Gamma \left( x \right) \cdot \Gamma \left( y \right)}}<br />
{{\Gamma \left( {x + y} \right)}}<br />
(and some values of <br />
{\Gamma \left( x \right)}<br />
; <br />
{\Gamma '\left( x \right)}<br />
and <br />
{\Gamma ''\left( x \right)}<br />
)

    That's not a nice computation but it still works
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  5. #20
    Super Member wingless's Avatar
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    Quote Originally Posted by Isomorphism View Post
    Here is the latest one:

    \boxed{I = \int_0^1 \int_0^x \frac{\ln (1 - y)}{y(y-1)}  \, dy \, dx}

    I invented it after seeing a recent problem

    \int_0^1 \int_0^x \frac{\ln (1 - y)}{y(y-1)}   ~ dy ~ dx

    \int_0^1 \int_y^1 \frac{\ln (1 - y)}{y(y-1)}  ~dx~dy

    \int_0^1 x\frac{\ln (1 - y)}{y(y-1)} \bigg{|}_y^1~dy

    \int_0^1 \frac{\ln (1-y)}{y(y-1)}-\frac{y\ln(1-y)}{y(y-1)} ~dy

    - \int_0^1 \frac{\ln (1-y)}{y}~dy

    \ln (1-y) = -\sum_{k=1}^{\infty} \frac{y^k}{k} = - \left (  y + \frac{y^2}{2}+\frac{y^3}{3}+\dots \right)

    \frac{\ln (1-y)}{y} = -\sum_{k=1}^{\infty} \frac{y^{k-1}}{k} = - \left (  1 + \frac{y}{2}+\frac{y^2}{3}+\dots \right )

    - \int_0^1 \frac{\ln (1-y)}{y}~dy = \sum_{k=1}^{\infty} \frac{y^k}{k^2}  \bigg{|}_0^1

    \sum_{k=1}^{\infty} \frac{1}{k^2}= \frac{\pi^2}{6}
    Last edited by wingless; July 4th 2008 at 06:15 AM.
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  6. #21
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    Quote Originally Posted by Isomorphism View Post
    Here is the latest update of "Definitely Integral"

    Compute I:

    \boxed{I = \int_0^{\pi/4}\ln(\sin x) \cdot \ln(\cos x)\,dx}
    First, consider \ln(1 + e^{2ix}) = \ln(2\cos x e^{ix})

    Comparing real parts you get \ln(2\cos x) = \frac{\cos 2x}{1} - \frac{\cos 4x}{2} + \frac{\cos 6x}{6} - \cdots

    \Rightarrow \ln(2 \sin x) = -\left(\frac{\cos 2x}{1} + \frac{\cos 4x}{2} + \frac{\cos 6x}{3} + \cdots \right)

    (I cannot rigorously justify these identities, but they seem to work at least for the range 0 \leq x \leq \frac{\pi}{2})

    Note that \int^{\frac{\pi}{4}}_{0} \ln(\sin x)\ln(\cos x) \;\mathrm{d}x = \frac{1}{2} \int^{\frac{\pi}{2}}_{0} \ln(\sin x)\ln(\cos x) \; \mathrm{d}x

    Also, if m \not= n, then \int^{\frac{\pi}{2}}_{0} \cos 2mx \cos 2nx \; \mathrm{d}x = \frac{1}{2} \int^{\frac{\pi}{2}}_{0} \cos 2(m+n)x + \cos 2(m-n) x \; \mathrm{d}x = 0

    So \int^{\frac{\pi}{2}}_{0} \ln(2\sin x)\ln(2\cos x) \; \mathrm{d}x = - \int^{\frac{\pi}{2}}_{0} \frac{\cos^{2}2x}{1^{2}} - \frac{\cos^{2}4x}{2^{2}} + \frac{\cos^{2}6x}{3^{2}} - \cdots \; \mathrm{d}x

    \int^{\frac{\pi}{2}}_{0} \cos^{2}2nx \; \mathrm{d}x = \int^{\frac{\pi}{2}}_{0} \sin^{2}2nx \; \mathrm{d}x = \frac{\pi}{4}

    \Rightarrow \int^{\frac{\pi}{2}}_{0} \ln(2\cos x)\ln(2\sin x) \; \mathrm{d}x = - \frac{\pi}{4}(\frac{1}{1^{2}} - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \cdots ) = - \frac{\pi^{3}}{48}

    \ln(\sin x)\ln(\cos x) = \ln(2 \sin x)\ln(2 \cos x) - \ln2\left( \ln(\sin x) + \ln(\cos x)\right) - (\ln2)^{2}

    It is a well-known result that \int^{\frac{\pi}{2}}_{0} \ln(\cos x) \; \mathrm{d}x = \int^{\frac{\pi}{2}}_{0} \ln(\sin x) \; \mathrm{d}x = -\frac{\pi}{2} \ln 2

    So \int^{\frac{\pi}{2}}_{0} \ln(\sin x)\ln(\cos x) \; \mathrm{d}x = -\frac{\pi^{3}}{48} + \pi (\ln2)^{2} - \frac{\pi}{2}(\ln 2)^{2} = \frac{\pi }{2}(\ln 2)^{2} - \frac{\pi^{3}}{48}

    \boxed{\Rightarrow \int^{\frac{\pi}{4}}_{0} \ln(\sin x)\ln(\cos x) \; \mathrm{d}x = \frac{\pi}{4} (\ln 2)^{2} - \frac{\pi^{3}}{96}}

    Awesome integral!
    Last edited by Dystopia; July 4th 2008 at 06:21 AM.
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  7. #22
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    Quote Originally Posted by wingless View Post
    \int_0^1 x\frac{\ln (1 - y)}{y(y-1)} \bigg{|}_1^y~dy
    Good job but I think you mean \bigg|_y^1.

    - \int_0^1 \frac{\ln (1-y)}{y}~dy = \sum_{k=1}^{\infty} \frac{y^k}{k^2}  \bigg{|}_0^1
    You are just missing an integral sign.
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  8. #23
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    Quote Originally Posted by Isomorphism View Post
    \int_0^\infty \frac{dx}{1+x^4} can be solved without complex numbers. I used only pure substitution and algebraic tricks
    Hmm, did you use the substitution x = e^{u}? That comes out fairly nicely, but there might be an easier way.

    Let I = \int^{\infty}_{0} \frac{1}{1 + x^{4}} \; \mathrm{d}x

    Let x = e^{u}

    I = \int^{\infty}_{-\infty} \frac{e^{u}}{1 + e^{4u}} \; \mathrm{d}x = \int^{\infty}_{-\infty} \frac{e^{-u}}{e^{2u} + e^{-2u}} \; \mathrm{d}x

    Let u = -t:

    I = \int^{\infty}_{-\infty} \frac{e^{t}}{e^{2t} + e^{-2t}} \;\mathrm{d}t

    \Rightarrow 2I = \int^{\infty}_{-\infty} \frac{\cosh t}{\cosh 2t} \; \mathrm{d}t

     = \int^{\infty}_{-\infty} \frac{\cosh t}{1 + 2\sinh^{2}t} \; \mathrm{d}t = \frac{\sqrt{2}}{2} \left[\tan^{-1}(\sinh t \cdot{\sqrt{2}})\right]^{\infty}_{-\infty} = \frac{\pi \sqrt{2}}{2}

    \boxed{\Rightarrow \int^{\infty}_{0} \frac{1}{1 + x^{4}} \; \mathrm{d}x = \frac{\pi \sqrt{2}}{4}}
    Last edited by Dystopia; July 4th 2008 at 06:36 AM.
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  9. #24
    Super Member wingless's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Good job but I think you mean \bigg|_y^1.
    True, typo fixed.

    Quote Originally Posted by ThePerfectHacker View Post
    You are just missing an integral sign.
    No, RHS is already integrated, otherwise it would be \int_0^1 \sum_{k=1}^{\infty}\frac{y^{k-1}}{k}~dy.
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  10. #25
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    Here is another:

    \int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}dxdy

    that works out to \sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2}  }{6}
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  11. #26
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    Quote Originally Posted by galactus View Post
    Here is another:
    \int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}dxdy
    \frac{1}{1-xy} = \sum_{n=0}^{\infty}(xy)^n.

    Thus, \int_0^1 \int_0^1 \sum_{n=0}^{\infty} x^n y^n dx dy = \int_0^1 \sum_{n=0}^{\infty}\frac{1}{n+1}y^n dy = \sum_{n=0}^{\infty}\frac{1}{(n+1)^2} = \frac{\pi^2}{6}.
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