Integral $int_{-infty}^{infty} frac{sinh(x)}{x [a+cosh(x)]^2}dx$












0












$begingroup$


I have difficulties with calculating the following integral:




$$I(a)=int_{-infty}^{infty} frac{sinh(x)}{x [a+cosh(x)]^2} mathrm dx~~~~~~~,text{where } a>1$$




For the case with $a=1$ the solution is $-28 zeta'(-2)$.






integration definite-integrals riemann-zeta hyperbolic-functions







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$endgroup$








  • 2




    $begingroup$
    Expand $frac{sinh(x)}{(a+cosh(x))^2} = sum_{k=1}^infty c_k(a) e^{-kx}$ then $int_0^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$ and $int_{-infty}^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = 2lim_{s to 0} sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$. Things like $zeta'(-2)$ appear from the functional equation, or the residue theorem.
    $endgroup$
    – reuns
    Jan 29 at 21:14








  • 2




    $begingroup$
    What have you tried? Where did you struggle? Please show some effort you made and include it within your post with an edit :)
    $endgroup$
    – mrtaurho
    Jan 29 at 21:15










  • $begingroup$
    Actually, I've tried an expansion but a very crude one. The solution by @reuns looks impressive.
    $endgroup$
    – Gregory Rut
    Jan 29 at 21:17
















0












$begingroup$


I have difficulties with calculating the following integral:




$$I(a)=int_{-infty}^{infty} frac{sinh(x)}{x [a+cosh(x)]^2} mathrm dx~~~~~~~,text{where } a>1$$




For the case with $a=1$ the solution is $-28 zeta'(-2)$.






integration definite-integrals riemann-zeta hyperbolic-functions







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Expand $frac{sinh(x)}{(a+cosh(x))^2} = sum_{k=1}^infty c_k(a) e^{-kx}$ then $int_0^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$ and $int_{-infty}^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = 2lim_{s to 0} sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$. Things like $zeta'(-2)$ appear from the functional equation, or the residue theorem.
    $endgroup$
    – reuns
    Jan 29 at 21:14








  • 2




    $begingroup$
    What have you tried? Where did you struggle? Please show some effort you made and include it within your post with an edit :)
    $endgroup$
    – mrtaurho
    Jan 29 at 21:15










  • $begingroup$
    Actually, I've tried an expansion but a very crude one. The solution by @reuns looks impressive.
    $endgroup$
    – Gregory Rut
    Jan 29 at 21:17














0












0








0





$begingroup$


I have difficulties with calculating the following integral:




$$I(a)=int_{-infty}^{infty} frac{sinh(x)}{x [a+cosh(x)]^2} mathrm dx~~~~~~~,text{where } a>1$$




For the case with $a=1$ the solution is $-28 zeta'(-2)$.






integration definite-integrals riemann-zeta hyperbolic-functions







share|cite|improve this question











$endgroup$




I have difficulties with calculating the following integral:




$$I(a)=int_{-infty}^{infty} frac{sinh(x)}{x [a+cosh(x)]^2} mathrm dx~~~~~~~,text{where } a>1$$




For the case with $a=1$ the solution is $-28 zeta'(-2)$.






integration definite-integrals riemann-zeta hyperbolic-functions




integration definite-integrals riemann-zeta hyperbolic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 0:42









Zacky

7,81511062




7,81511062










asked Jan 29 at 21:02









Gregory RutGregory Rut

1092




1092








  • 2




    $begingroup$
    Expand $frac{sinh(x)}{(a+cosh(x))^2} = sum_{k=1}^infty c_k(a) e^{-kx}$ then $int_0^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$ and $int_{-infty}^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = 2lim_{s to 0} sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$. Things like $zeta'(-2)$ appear from the functional equation, or the residue theorem.
    $endgroup$
    – reuns
    Jan 29 at 21:14








  • 2




    $begingroup$
    What have you tried? Where did you struggle? Please show some effort you made and include it within your post with an edit :)
    $endgroup$
    – mrtaurho
    Jan 29 at 21:15










  • $begingroup$
    Actually, I've tried an expansion but a very crude one. The solution by @reuns looks impressive.
    $endgroup$
    – Gregory Rut
    Jan 29 at 21:17














  • 2




    $begingroup$
    Expand $frac{sinh(x)}{(a+cosh(x))^2} = sum_{k=1}^infty c_k(a) e^{-kx}$ then $int_0^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$ and $int_{-infty}^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = 2lim_{s to 0} sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$. Things like $zeta'(-2)$ appear from the functional equation, or the residue theorem.
    $endgroup$
    – reuns
    Jan 29 at 21:14








  • 2




    $begingroup$
    What have you tried? Where did you struggle? Please show some effort you made and include it within your post with an edit :)
    $endgroup$
    – mrtaurho
    Jan 29 at 21:15










  • $begingroup$
    Actually, I've tried an expansion but a very crude one. The solution by @reuns looks impressive.
    $endgroup$
    – Gregory Rut
    Jan 29 at 21:17








2




2




$begingroup$
Expand $frac{sinh(x)}{(a+cosh(x))^2} = sum_{k=1}^infty c_k(a) e^{-kx}$ then $int_0^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$ and $int_{-infty}^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = 2lim_{s to 0} sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$. Things like $zeta'(-2)$ appear from the functional equation, or the residue theorem.
$endgroup$
– reuns
Jan 29 at 21:14






$begingroup$
Expand $frac{sinh(x)}{(a+cosh(x))^2} = sum_{k=1}^infty c_k(a) e^{-kx}$ then $int_0^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$ and $int_{-infty}^infty x^{s-1} frac{sinh(x)}{(a+cosh(x))^2} dx = 2lim_{s to 0} sum_{k=1}^infty c_k(a) k^{-s} Gamma(s)$. Things like $zeta'(-2)$ appear from the functional equation, or the residue theorem.
$endgroup$
– reuns
Jan 29 at 21:14






2




2




$begingroup$
What have you tried? Where did you struggle? Please show some effort you made and include it within your post with an edit :)
$endgroup$
– mrtaurho
Jan 29 at 21:15




$begingroup$
What have you tried? Where did you struggle? Please show some effort you made and include it within your post with an edit :)
$endgroup$
– mrtaurho
Jan 29 at 21:15












$begingroup$
Actually, I've tried an expansion but a very crude one. The solution by @reuns looks impressive.
$endgroup$
– Gregory Rut
Jan 29 at 21:17




$begingroup$
Actually, I've tried an expansion but a very crude one. The solution by @reuns looks impressive.
$endgroup$
– Gregory Rut
Jan 29 at 21:17










1 Answer
1






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$begingroup$

I will do the case $a=1$ in order to give some insight:$$I=int_{-infty}^{infty} frac{sinh(x)}{x (1+cosh(x))^2}dxoverset{x=ln t}=2int_0^infty frac{t-1}{(t+1)^3}frac{dt}{ln t}$$
We can consider the following integral and perform Feynman's trick:
$$I(n)=2int_0^infty frac{t^{n-1}-1}{(t+1)^3}frac{dt}{ln t}$$
$$Rightarrow frac{d}{dn}I(n)=2int_0^infty frac{t^{n-1}}{(t+1)^3}dt=2B(n,3-n)=Gamma(n)Gamma(3-n)$$
$$=(2-n)(1-n)Gamma(n)Gamma(1-n)=pifrac{(1-n)(2-n)}{sin(pi n)}$$
$$I(1)=0 Rightarrow I(2)-I(1)=I =piint_1^2 frac{(1-n)(2-n)}{sin(pi n)}dn$$
$$ overset{n-1=x}=boxed{pi int_0^1 frac{x(1-x)}{sin(pi x)}dx = frac{7zeta(3)}{pi^2 }=-28zeta'(-2)}$$
The last integral can be found here.



By the same idea one gets: $$I(a)=int_{-infty}^infty frac{sinh x}{x(a+cosh x))^2}dx=2int_0^infty frac{x^2-1}{(x^2+2ax+1)^2}frac{dx}{ln x}$$
Now consider the same parameter take a derivative with respect to it, apply a Mellin transform, and try to carry on.






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    1 Answer
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    active

    oldest

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    4












    $begingroup$

    I will do the case $a=1$ in order to give some insight:$$I=int_{-infty}^{infty} frac{sinh(x)}{x (1+cosh(x))^2}dxoverset{x=ln t}=2int_0^infty frac{t-1}{(t+1)^3}frac{dt}{ln t}$$
    We can consider the following integral and perform Feynman's trick:
    $$I(n)=2int_0^infty frac{t^{n-1}-1}{(t+1)^3}frac{dt}{ln t}$$
    $$Rightarrow frac{d}{dn}I(n)=2int_0^infty frac{t^{n-1}}{(t+1)^3}dt=2B(n,3-n)=Gamma(n)Gamma(3-n)$$
    $$=(2-n)(1-n)Gamma(n)Gamma(1-n)=pifrac{(1-n)(2-n)}{sin(pi n)}$$
    $$I(1)=0 Rightarrow I(2)-I(1)=I =piint_1^2 frac{(1-n)(2-n)}{sin(pi n)}dn$$
    $$ overset{n-1=x}=boxed{pi int_0^1 frac{x(1-x)}{sin(pi x)}dx = frac{7zeta(3)}{pi^2 }=-28zeta'(-2)}$$
    The last integral can be found here.



    By the same idea one gets: $$I(a)=int_{-infty}^infty frac{sinh x}{x(a+cosh x))^2}dx=2int_0^infty frac{x^2-1}{(x^2+2ax+1)^2}frac{dx}{ln x}$$
    Now consider the same parameter take a derivative with respect to it, apply a Mellin transform, and try to carry on.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      I will do the case $a=1$ in order to give some insight:$$I=int_{-infty}^{infty} frac{sinh(x)}{x (1+cosh(x))^2}dxoverset{x=ln t}=2int_0^infty frac{t-1}{(t+1)^3}frac{dt}{ln t}$$
      We can consider the following integral and perform Feynman's trick:
      $$I(n)=2int_0^infty frac{t^{n-1}-1}{(t+1)^3}frac{dt}{ln t}$$
      $$Rightarrow frac{d}{dn}I(n)=2int_0^infty frac{t^{n-1}}{(t+1)^3}dt=2B(n,3-n)=Gamma(n)Gamma(3-n)$$
      $$=(2-n)(1-n)Gamma(n)Gamma(1-n)=pifrac{(1-n)(2-n)}{sin(pi n)}$$
      $$I(1)=0 Rightarrow I(2)-I(1)=I =piint_1^2 frac{(1-n)(2-n)}{sin(pi n)}dn$$
      $$ overset{n-1=x}=boxed{pi int_0^1 frac{x(1-x)}{sin(pi x)}dx = frac{7zeta(3)}{pi^2 }=-28zeta'(-2)}$$
      The last integral can be found here.



      By the same idea one gets: $$I(a)=int_{-infty}^infty frac{sinh x}{x(a+cosh x))^2}dx=2int_0^infty frac{x^2-1}{(x^2+2ax+1)^2}frac{dx}{ln x}$$
      Now consider the same parameter take a derivative with respect to it, apply a Mellin transform, and try to carry on.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        I will do the case $a=1$ in order to give some insight:$$I=int_{-infty}^{infty} frac{sinh(x)}{x (1+cosh(x))^2}dxoverset{x=ln t}=2int_0^infty frac{t-1}{(t+1)^3}frac{dt}{ln t}$$
        We can consider the following integral and perform Feynman's trick:
        $$I(n)=2int_0^infty frac{t^{n-1}-1}{(t+1)^3}frac{dt}{ln t}$$
        $$Rightarrow frac{d}{dn}I(n)=2int_0^infty frac{t^{n-1}}{(t+1)^3}dt=2B(n,3-n)=Gamma(n)Gamma(3-n)$$
        $$=(2-n)(1-n)Gamma(n)Gamma(1-n)=pifrac{(1-n)(2-n)}{sin(pi n)}$$
        $$I(1)=0 Rightarrow I(2)-I(1)=I =piint_1^2 frac{(1-n)(2-n)}{sin(pi n)}dn$$
        $$ overset{n-1=x}=boxed{pi int_0^1 frac{x(1-x)}{sin(pi x)}dx = frac{7zeta(3)}{pi^2 }=-28zeta'(-2)}$$
        The last integral can be found here.



        By the same idea one gets: $$I(a)=int_{-infty}^infty frac{sinh x}{x(a+cosh x))^2}dx=2int_0^infty frac{x^2-1}{(x^2+2ax+1)^2}frac{dx}{ln x}$$
        Now consider the same parameter take a derivative with respect to it, apply a Mellin transform, and try to carry on.






        share|cite|improve this answer











        $endgroup$



        I will do the case $a=1$ in order to give some insight:$$I=int_{-infty}^{infty} frac{sinh(x)}{x (1+cosh(x))^2}dxoverset{x=ln t}=2int_0^infty frac{t-1}{(t+1)^3}frac{dt}{ln t}$$
        We can consider the following integral and perform Feynman's trick:
        $$I(n)=2int_0^infty frac{t^{n-1}-1}{(t+1)^3}frac{dt}{ln t}$$
        $$Rightarrow frac{d}{dn}I(n)=2int_0^infty frac{t^{n-1}}{(t+1)^3}dt=2B(n,3-n)=Gamma(n)Gamma(3-n)$$
        $$=(2-n)(1-n)Gamma(n)Gamma(1-n)=pifrac{(1-n)(2-n)}{sin(pi n)}$$
        $$I(1)=0 Rightarrow I(2)-I(1)=I =piint_1^2 frac{(1-n)(2-n)}{sin(pi n)}dn$$
        $$ overset{n-1=x}=boxed{pi int_0^1 frac{x(1-x)}{sin(pi x)}dx = frac{7zeta(3)}{pi^2 }=-28zeta'(-2)}$$
        The last integral can be found here.



        By the same idea one gets: $$I(a)=int_{-infty}^infty frac{sinh x}{x(a+cosh x))^2}dx=2int_0^infty frac{x^2-1}{(x^2+2ax+1)^2}frac{dx}{ln x}$$
        Now consider the same parameter take a derivative with respect to it, apply a Mellin transform, and try to carry on.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 30 at 15:44

























        answered Jan 30 at 0:37









        ZackyZacky

        7,81511062




        7,81511062






























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