Looking for another way to calculate the integral $iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A$












3












$begingroup$


Here, I have a little unpleasant way to calculate the following double integral



$$iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A$$



where $D$ is the square area $D={(x,y)inmathbb{R}^2: 0 le x le pi/2, 0 le y le pi/2}$



my attempt:



with the symmetry of the area we know



$$I=iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A=iint_{D}{sin(y)e^{sin(x)sin(y)}}text{d}A$$



thus



$$I=frac1{2}iint_{D}{(sin(x)+sin(y))e^{sin(x)sin(y)}}text{d}A$$



with the substitution $u=sin(x)$ and $v=sin(y)$, we have



$$I=frac1{2}iint_{D^{*}}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A$$



and now the integral area is $D^{*}={(u,v)inmathbb{R}^2: 0 le u le 1, 0 le v le 1}$



notice that the function $frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}$ also holds a symmetric form, thus



$$I=iint_{D^{*}cap(ule v)}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A=iint_{D^{*}cap(uge v)}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A$$



so I can only foucs on area $D^{*}cap(uge v)$, set another substitution



$$u+v=alpha$$



$$uv=beta$$



under the condition $uge v$, the determinant of Jacobian matrix writes



$$|J(alpha,beta)|=frac1{sqrt{alpha^2-4beta}}$$



and



$$frac1{sqrt{(1-u^2)(1-v^2)}}=frac1{sqrt{(1+beta)^2-alpha^2}}$$



the integral area change to ${(alpha,beta)inmathbb{R}^2: 2sqrt{beta} le alpha le 1+beta, 0 le beta le 1}$, so



$$begin{align}
I&=int_{0}^{1}int_{2sqrt{beta}}^{1+beta}{frac{alpha}{sqrt{((1+beta)^2-alpha^2)(alpha^2-4beta)}}e^{beta}}text{d}alpha text{d}beta
\&=int_{0}^{1}left(-tan^{-1}sqrt{frac{(1+beta)^2-alpha^2}{alpha^2-4beta}}right)biggr|_{alpha=2sqrt{beta}}^{1+beta} e^{beta}text{d}beta
\&=frac{pi}{2}(e-1)
end{align}$$



obviously, I choose an inconvenient way in the second half of this calculation, but I still can not come up with another good method.



thanks in advance for any suggestion!






calculus multivariable-calculus







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Here, I have a little unpleasant way to calculate the following double integral



    $$iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A$$



    where $D$ is the square area $D={(x,y)inmathbb{R}^2: 0 le x le pi/2, 0 le y le pi/2}$



    my attempt:



    with the symmetry of the area we know



    $$I=iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A=iint_{D}{sin(y)e^{sin(x)sin(y)}}text{d}A$$



    thus



    $$I=frac1{2}iint_{D}{(sin(x)+sin(y))e^{sin(x)sin(y)}}text{d}A$$



    with the substitution $u=sin(x)$ and $v=sin(y)$, we have



    $$I=frac1{2}iint_{D^{*}}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A$$



    and now the integral area is $D^{*}={(u,v)inmathbb{R}^2: 0 le u le 1, 0 le v le 1}$



    notice that the function $frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}$ also holds a symmetric form, thus



    $$I=iint_{D^{*}cap(ule v)}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A=iint_{D^{*}cap(uge v)}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A$$



    so I can only foucs on area $D^{*}cap(uge v)$, set another substitution



    $$u+v=alpha$$



    $$uv=beta$$



    under the condition $uge v$, the determinant of Jacobian matrix writes



    $$|J(alpha,beta)|=frac1{sqrt{alpha^2-4beta}}$$



    and



    $$frac1{sqrt{(1-u^2)(1-v^2)}}=frac1{sqrt{(1+beta)^2-alpha^2}}$$



    the integral area change to ${(alpha,beta)inmathbb{R}^2: 2sqrt{beta} le alpha le 1+beta, 0 le beta le 1}$, so



    $$begin{align}
    I&=int_{0}^{1}int_{2sqrt{beta}}^{1+beta}{frac{alpha}{sqrt{((1+beta)^2-alpha^2)(alpha^2-4beta)}}e^{beta}}text{d}alpha text{d}beta
    \&=int_{0}^{1}left(-tan^{-1}sqrt{frac{(1+beta)^2-alpha^2}{alpha^2-4beta}}right)biggr|_{alpha=2sqrt{beta}}^{1+beta} e^{beta}text{d}beta
    \&=frac{pi}{2}(e-1)
    end{align}$$



    obviously, I choose an inconvenient way in the second half of this calculation, but I still can not come up with another good method.



    thanks in advance for any suggestion!






    calculus multivariable-calculus







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Here, I have a little unpleasant way to calculate the following double integral



      $$iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A$$



      where $D$ is the square area $D={(x,y)inmathbb{R}^2: 0 le x le pi/2, 0 le y le pi/2}$



      my attempt:



      with the symmetry of the area we know



      $$I=iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A=iint_{D}{sin(y)e^{sin(x)sin(y)}}text{d}A$$



      thus



      $$I=frac1{2}iint_{D}{(sin(x)+sin(y))e^{sin(x)sin(y)}}text{d}A$$



      with the substitution $u=sin(x)$ and $v=sin(y)$, we have



      $$I=frac1{2}iint_{D^{*}}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A$$



      and now the integral area is $D^{*}={(u,v)inmathbb{R}^2: 0 le u le 1, 0 le v le 1}$



      notice that the function $frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}$ also holds a symmetric form, thus



      $$I=iint_{D^{*}cap(ule v)}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A=iint_{D^{*}cap(uge v)}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A$$



      so I can only foucs on area $D^{*}cap(uge v)$, set another substitution



      $$u+v=alpha$$



      $$uv=beta$$



      under the condition $uge v$, the determinant of Jacobian matrix writes



      $$|J(alpha,beta)|=frac1{sqrt{alpha^2-4beta}}$$



      and



      $$frac1{sqrt{(1-u^2)(1-v^2)}}=frac1{sqrt{(1+beta)^2-alpha^2}}$$



      the integral area change to ${(alpha,beta)inmathbb{R}^2: 2sqrt{beta} le alpha le 1+beta, 0 le beta le 1}$, so



      $$begin{align}
      I&=int_{0}^{1}int_{2sqrt{beta}}^{1+beta}{frac{alpha}{sqrt{((1+beta)^2-alpha^2)(alpha^2-4beta)}}e^{beta}}text{d}alpha text{d}beta
      \&=int_{0}^{1}left(-tan^{-1}sqrt{frac{(1+beta)^2-alpha^2}{alpha^2-4beta}}right)biggr|_{alpha=2sqrt{beta}}^{1+beta} e^{beta}text{d}beta
      \&=frac{pi}{2}(e-1)
      end{align}$$



      obviously, I choose an inconvenient way in the second half of this calculation, but I still can not come up with another good method.



      thanks in advance for any suggestion!






      calculus multivariable-calculus







      share|cite|improve this question









      $endgroup$




      Here, I have a little unpleasant way to calculate the following double integral



      $$iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A$$



      where $D$ is the square area $D={(x,y)inmathbb{R}^2: 0 le x le pi/2, 0 le y le pi/2}$



      my attempt:



      with the symmetry of the area we know



      $$I=iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A=iint_{D}{sin(y)e^{sin(x)sin(y)}}text{d}A$$



      thus



      $$I=frac1{2}iint_{D}{(sin(x)+sin(y))e^{sin(x)sin(y)}}text{d}A$$



      with the substitution $u=sin(x)$ and $v=sin(y)$, we have



      $$I=frac1{2}iint_{D^{*}}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A$$



      and now the integral area is $D^{*}={(u,v)inmathbb{R}^2: 0 le u le 1, 0 le v le 1}$



      notice that the function $frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}$ also holds a symmetric form, thus



      $$I=iint_{D^{*}cap(ule v)}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A=iint_{D^{*}cap(uge v)}{frac{u+v}{sqrt{(1-u^2)(1-v^2)}}e^{uv}}text{d}A$$



      so I can only foucs on area $D^{*}cap(uge v)$, set another substitution



      $$u+v=alpha$$



      $$uv=beta$$



      under the condition $uge v$, the determinant of Jacobian matrix writes



      $$|J(alpha,beta)|=frac1{sqrt{alpha^2-4beta}}$$



      and



      $$frac1{sqrt{(1-u^2)(1-v^2)}}=frac1{sqrt{(1+beta)^2-alpha^2}}$$



      the integral area change to ${(alpha,beta)inmathbb{R}^2: 2sqrt{beta} le alpha le 1+beta, 0 le beta le 1}$, so



      $$begin{align}
      I&=int_{0}^{1}int_{2sqrt{beta}}^{1+beta}{frac{alpha}{sqrt{((1+beta)^2-alpha^2)(alpha^2-4beta)}}e^{beta}}text{d}alpha text{d}beta
      \&=int_{0}^{1}left(-tan^{-1}sqrt{frac{(1+beta)^2-alpha^2}{alpha^2-4beta}}right)biggr|_{alpha=2sqrt{beta}}^{1+beta} e^{beta}text{d}beta
      \&=frac{pi}{2}(e-1)
      end{align}$$



      obviously, I choose an inconvenient way in the second half of this calculation, but I still can not come up with another good method.



      thanks in advance for any suggestion!






      calculus multivariable-calculus




      calculus multivariable-calculus






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 9 '18 at 14:57









      NanayajitzukiNanayajitzuki

      3285




      3285






















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          $begingroup$

          Using the Taylor expansion of the exponential
          $$
          iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A=sum_{kgeq 0}frac{1}{k!}int_0^{pi/2}mathrm{d}x~(sin x)^{k+1}int_0^{pi/2}mathrm{d}y~(sin y)^{k}=sum_{kgeq 0}frac{1}{k!}I(k)I(k+1) ,
          $$
          where
          $$
          I(k)=int_0^{pi/2}mathrm{d}x~(sin x)^{k}=int_0^1mathrm{du}~u^k(1-u^2)^{-1/2}=frac{sqrt{pi}~Gamma left(frac{k+1}{2}right)}{2 Gamma left(frac{k}{2}+1right)} .
          $$
          Therefore
          $$
          frac{1}{k!}I(k)I(k+1)=frac{left(sqrt{pi } Gamma left(frac{k+1}{2}right)right) left(sqrt{pi } Gamma left(frac{k}{2}+1right)right)}{k! left(2 Gamma left(frac{k}{2}+1right)right) left(2 Gamma left(frac{k+3}{2}right)right)}=frac{pi }{2 Gamma (k+2)}
          $$
          after simplifications, which then leads to
          $$
          sum_{kgeq 0}frac{1}{k!}I(k)I(k+1)=frac{1}{2} (e-1) pi .
          $$






          share|cite|improve this answer









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            $begingroup$

            Using the Taylor expansion of the exponential
            $$
            iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A=sum_{kgeq 0}frac{1}{k!}int_0^{pi/2}mathrm{d}x~(sin x)^{k+1}int_0^{pi/2}mathrm{d}y~(sin y)^{k}=sum_{kgeq 0}frac{1}{k!}I(k)I(k+1) ,
            $$
            where
            $$
            I(k)=int_0^{pi/2}mathrm{d}x~(sin x)^{k}=int_0^1mathrm{du}~u^k(1-u^2)^{-1/2}=frac{sqrt{pi}~Gamma left(frac{k+1}{2}right)}{2 Gamma left(frac{k}{2}+1right)} .
            $$
            Therefore
            $$
            frac{1}{k!}I(k)I(k+1)=frac{left(sqrt{pi } Gamma left(frac{k+1}{2}right)right) left(sqrt{pi } Gamma left(frac{k}{2}+1right)right)}{k! left(2 Gamma left(frac{k}{2}+1right)right) left(2 Gamma left(frac{k+3}{2}right)right)}=frac{pi }{2 Gamma (k+2)}
            $$
            after simplifications, which then leads to
            $$
            sum_{kgeq 0}frac{1}{k!}I(k)I(k+1)=frac{1}{2} (e-1) pi .
            $$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Using the Taylor expansion of the exponential
              $$
              iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A=sum_{kgeq 0}frac{1}{k!}int_0^{pi/2}mathrm{d}x~(sin x)^{k+1}int_0^{pi/2}mathrm{d}y~(sin y)^{k}=sum_{kgeq 0}frac{1}{k!}I(k)I(k+1) ,
              $$
              where
              $$
              I(k)=int_0^{pi/2}mathrm{d}x~(sin x)^{k}=int_0^1mathrm{du}~u^k(1-u^2)^{-1/2}=frac{sqrt{pi}~Gamma left(frac{k+1}{2}right)}{2 Gamma left(frac{k}{2}+1right)} .
              $$
              Therefore
              $$
              frac{1}{k!}I(k)I(k+1)=frac{left(sqrt{pi } Gamma left(frac{k+1}{2}right)right) left(sqrt{pi } Gamma left(frac{k}{2}+1right)right)}{k! left(2 Gamma left(frac{k}{2}+1right)right) left(2 Gamma left(frac{k+3}{2}right)right)}=frac{pi }{2 Gamma (k+2)}
              $$
              after simplifications, which then leads to
              $$
              sum_{kgeq 0}frac{1}{k!}I(k)I(k+1)=frac{1}{2} (e-1) pi .
              $$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Using the Taylor expansion of the exponential
                $$
                iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A=sum_{kgeq 0}frac{1}{k!}int_0^{pi/2}mathrm{d}x~(sin x)^{k+1}int_0^{pi/2}mathrm{d}y~(sin y)^{k}=sum_{kgeq 0}frac{1}{k!}I(k)I(k+1) ,
                $$
                where
                $$
                I(k)=int_0^{pi/2}mathrm{d}x~(sin x)^{k}=int_0^1mathrm{du}~u^k(1-u^2)^{-1/2}=frac{sqrt{pi}~Gamma left(frac{k+1}{2}right)}{2 Gamma left(frac{k}{2}+1right)} .
                $$
                Therefore
                $$
                frac{1}{k!}I(k)I(k+1)=frac{left(sqrt{pi } Gamma left(frac{k+1}{2}right)right) left(sqrt{pi } Gamma left(frac{k}{2}+1right)right)}{k! left(2 Gamma left(frac{k}{2}+1right)right) left(2 Gamma left(frac{k+3}{2}right)right)}=frac{pi }{2 Gamma (k+2)}
                $$
                after simplifications, which then leads to
                $$
                sum_{kgeq 0}frac{1}{k!}I(k)I(k+1)=frac{1}{2} (e-1) pi .
                $$






                share|cite|improve this answer









                $endgroup$



                Using the Taylor expansion of the exponential
                $$
                iint_{D}{sin(x)e^{sin(x)sin(y)}}text{d}A=sum_{kgeq 0}frac{1}{k!}int_0^{pi/2}mathrm{d}x~(sin x)^{k+1}int_0^{pi/2}mathrm{d}y~(sin y)^{k}=sum_{kgeq 0}frac{1}{k!}I(k)I(k+1) ,
                $$
                where
                $$
                I(k)=int_0^{pi/2}mathrm{d}x~(sin x)^{k}=int_0^1mathrm{du}~u^k(1-u^2)^{-1/2}=frac{sqrt{pi}~Gamma left(frac{k+1}{2}right)}{2 Gamma left(frac{k}{2}+1right)} .
                $$
                Therefore
                $$
                frac{1}{k!}I(k)I(k+1)=frac{left(sqrt{pi } Gamma left(frac{k+1}{2}right)right) left(sqrt{pi } Gamma left(frac{k}{2}+1right)right)}{k! left(2 Gamma left(frac{k}{2}+1right)right) left(2 Gamma left(frac{k+3}{2}right)right)}=frac{pi }{2 Gamma (k+2)}
                $$
                after simplifications, which then leads to
                $$
                sum_{kgeq 0}frac{1}{k!}I(k)I(k+1)=frac{1}{2} (e-1) pi .
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 24 '18 at 15:17









                Pierpaolo VivoPierpaolo Vivo

                5,3812724




                5,3812724






























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