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Math Help - Today's calculation of integral #14

  1. #1
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    Today's calculation of integral #14

    Find \int_{0}^{\infty }\!{\int_{y}^{\infty }{\bigg\{\frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}\bigg\}\,dx}\,dy}.


    Clever substitutions are necessary to kill this double integral.
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  2. #2
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    Hint: let x=yz, and find another suitable substitution.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Krizalid View Post
    Find \int_{0}^{\infty }\!{\int_{y}^{\infty }{\bigg\{\frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}\bigg\}\,dx}\,dy}.


    Clever substitutions are necessary to kill this double integral.
    Quote Originally Posted by Krizalid View Post
    Hint: let x=yz, and find another suitable substitution.
    I'm trying to work it out, but I got stuck. I'll put what I have. Maybe you can help me out a little bit:

    \int_{0}^{\infty }\!{\int_{y}^{\infty }{\bigg\{\frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}\bigg\}\,dx}\,dy}.

    What I did first was rewrite the integrand with that hint you gave: let x=yz.

    \frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}
    =\frac{(yz-y)^{2}}{y^2z\sinh(x+y)}\cdot\ln\frac{yz+y}{yz-y}
    =\frac{(y(z-1))^{2}}{y^2z\sinh(y(z+1))}\cdot\ln\frac{y(z+1)}{y  (z-1)}
    =\frac{y^{2}(z-1)^{2}}{y^2z\cdot e^{y}\sinh(z+1)}\cdot\ln\frac{z+1}{z-1}
    =\frac{(z-1)^{2}}{z\cdot e^{y}\sinh(z+1)}\cdot\ln\frac{z+1}{z-1}.

    Here, I made another substitution: let y=ln\frac{z+1}{z-1}.

    Now I got:

    \frac{{(z-1)^{2}y}}{z\cdot e^{y}\sinh(z+1)}.

    I have question on converting the limits of integration, and getting the correct differentials (dz and dy)

    From the first substitution x=yz, I get dx in terms of dy and dz:

    \,dx=y\,dz+z\,dy

    From the second substitution y=ln\frac{z+1}{z-1}, I get dy:

    \,dy=\frac{z-1}{z+1}\cdot \frac{(z-1)(1)-(z+1)(1)}{(z-1)^{2}}dz
    \,dy=\frac{z-1}{z+1}\cdot \frac{-2}{(z-1)^{2}}dz
    \,dy=-\frac{2}{(z+1)(z-1)}dz.

    Substituting this into the equation for dx, I get:

    \,dx=y\,dz+z{\left(-\frac{2}{(z+1)(z-1)}dz\right)}
    \,dx=y\,dz-\frac{2z}{(z+1)(z-1)}dz.

    Substituting, into the integral, I get:

    \int_{0}^{\infty}\int_{?}^{?}{\left(\frac{{(z-1)^{2}y}}{z\cdot e^{y}sinh(z+1)}\right)}{\left(y\,dz-\frac{2z}{(z+1)(z-1)}\,dz\right)}\,dy
    =\int_{0}^{\infty}\int_{?}^{?}{\left(\frac{{(z-1)^{2}y}}{z\cdot e^{y}sinh(z+1)}\right)}{\left(y\,dz\,dy-\frac{2z}{(z+1)(z-1)}\,dz\,dy\right)}
    =\int_{0}^{\infty}\int_{?}^{?}{\left(\frac{{(z-1)^{2}y^{2}}}{z\cdot e^{y}sinh(z+1)}\right)}\,dz\,dy-\int_{0}^{\infty}\int_{?}^{?}{\left(\frac{2y}{e^{2  y}sinh(z+1)}\right)}\,dz\,dy.

    I'm still stuck. I made a couple changes and cleaned it up a little bit. How do I determine the limits for dz?

    ...at least I gave it a shot!
    Last edited by Chris L T521; May 9th 2008 at 08:00 PM. Reason: Figured something out...but still need help on limits of integration...
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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
    Hint: let x=yz, and find another suitable substitution.
    Now who would think of something as crazy as that?!
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  5. #5
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    Before to make the another substitution, we require another previous one.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Krizalid View Post
    Before to make the another substitution, we require another previous one.
    Another substitution? I have to admit, this is a very good problem...a bit challenging, though.

    I'm thinking I need to make a substitution for the sinh(z+1) term???
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  7. #7
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    Ahhh, for the inner integral we require the substitution x=yz. ( z is the new variable, so we use simple differentiation here and not product rule. After this substitution you may have to invoke another one, and then use... figure out. )
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Krizalid View Post
    Ahhh, for the inner integral we require the substitution x=yz. ( z is the new variable, so we use simple differentiation here and not product rule. After this substitution you may have to invoke another one, and then use... figure out. )
    So are you saying that after we differentiate x, we should get: \,dx=y\,dz?
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  9. #9
    Math Engineering Student
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    I'm pretty tired to post a full solution so I'll post the main steps to tackle this.

    Quote Originally Posted by Krizalid View Post

    Find \int_{0}^{\infty }\!{\int_{y}^{\infty }{\bigg\{\frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}\bigg\}\,dx}\,dy}.
    Substitute x=yz for the inner integral and reverse integration order, then the original one becomes \int_{1}^{\infty }\!\!{\int_{0}^{\infty }{\left\{ \frac{(z-1)^{2}}{z}\cdot \ln \frac{(z+1)}{(z-1)} \right\}\left\{ \frac{y}{\sinh \left( y(z+1) \right)} \right\}\,dy}\,dz}. Now let \varphi =y(z+1) and the double integral becomes \int_{1}^{\infty }\!\!{\int_{0}^{\infty }{\Bigg\{ \frac{(z-1)^{2}}{z(z+1)^{2}}\cdot \ln \frac{(z+1)}{(z-1)} \Bigg\}\Bigg\{ \frac{\varphi }{\sinh \varphi } \Bigg\}\,d\varphi }\,dz}.

    Consider f(\alpha )=\int_{0}^{\infty }{\frac{xe^{-\alpha x}}{1-e^{-2x}}\,dx},\,\forall\,\alpha>0, then f(\alpha )=\sum\limits_{k\,=\,0}^{\infty }{\Bigg\{ \int_{0}^{\infty }{xe^{-(2k+\alpha )x}\,dx} \Bigg\}}=\sum\limits_{k\,=\,0}^{\infty }{\frac{1}{(2k+\alpha )^{2}}}. From here we have \int_{0}^{\infty }{\frac{\varphi }{\sinh\varphi }\,d\varphi }=2\int_{0}^{\infty }{\frac{e^{-\varphi }}{1-e^{-2\varphi }}\,d\varphi }=\frac{\pi ^{2}}{4}.

    Finally, for the remaining integral substitute \mu =\ln \dfrac{z+1}{z-1}, and we happily get

    \frac{\pi ^{2}}{2}\int_{0}^{\infty }{\frac{\mu e^{-3\mu }}{1-e^{-2\mu }}\,d\mu}<br />
  =\frac{\pi ^{2}}{2}\sum\limits_{k\,=\,0}^{\infty }{\frac{1}{(2k+3)^{2}}}<br />
 =\frac{\pi ^{2}}{2}\sum\limits_{k\,=\,1}^{\infty }{\frac{1}{(2k+1)^{2}}}, hence the original double integral equals 3\zeta (2)\Bigg\{ \frac{3}{4}\zeta (2)-1 \Bigg\}.
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