The integral $int_{-infty}^{infty}frac{1}{R^2+z^2}e^{-alpha sqrt{R^2+z^2}}mathrm{d}z$












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I need to evaluate the following integral (for a physical application):
$$I=int_{-infty}^{infty}frac{1}{R^2+z^2}e^{-alpha sqrt{R^2+z^2}}mathrm{d}z, $$
where $alpha>0$ and $R>0$.



I tried putting it in Mathematica, but it failed. I thought about using the residue theorem. I thought that the only pole inside the upper half of the complex plane should be at $z=iR$ (because $R^2+z^2=(z-iR)(z+iR)$). But when evaluating the residue at the simple pole $z=iR$, I am left with an answer independent of $alpha$, which does not make sense physically.










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  • $begingroup$
    You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
    $endgroup$
    – Zachary
    Dec 10 '18 at 20:25










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    I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
    $endgroup$
    – projectilemotion
    Dec 10 '18 at 20:57


















3












$begingroup$


I need to evaluate the following integral (for a physical application):
$$I=int_{-infty}^{infty}frac{1}{R^2+z^2}e^{-alpha sqrt{R^2+z^2}}mathrm{d}z, $$
where $alpha>0$ and $R>0$.



I tried putting it in Mathematica, but it failed. I thought about using the residue theorem. I thought that the only pole inside the upper half of the complex plane should be at $z=iR$ (because $R^2+z^2=(z-iR)(z+iR)$). But when evaluating the residue at the simple pole $z=iR$, I am left with an answer independent of $alpha$, which does not make sense physically.










share|cite|improve this question









$endgroup$












  • $begingroup$
    You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
    $endgroup$
    – Zachary
    Dec 10 '18 at 20:25










  • $begingroup$
    I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
    $endgroup$
    – projectilemotion
    Dec 10 '18 at 20:57
















3












3








3


1



$begingroup$


I need to evaluate the following integral (for a physical application):
$$I=int_{-infty}^{infty}frac{1}{R^2+z^2}e^{-alpha sqrt{R^2+z^2}}mathrm{d}z, $$
where $alpha>0$ and $R>0$.



I tried putting it in Mathematica, but it failed. I thought about using the residue theorem. I thought that the only pole inside the upper half of the complex plane should be at $z=iR$ (because $R^2+z^2=(z-iR)(z+iR)$). But when evaluating the residue at the simple pole $z=iR$, I am left with an answer independent of $alpha$, which does not make sense physically.










share|cite|improve this question









$endgroup$




I need to evaluate the following integral (for a physical application):
$$I=int_{-infty}^{infty}frac{1}{R^2+z^2}e^{-alpha sqrt{R^2+z^2}}mathrm{d}z, $$
where $alpha>0$ and $R>0$.



I tried putting it in Mathematica, but it failed. I thought about using the residue theorem. I thought that the only pole inside the upper half of the complex plane should be at $z=iR$ (because $R^2+z^2=(z-iR)(z+iR)$). But when evaluating the residue at the simple pole $z=iR$, I am left with an answer independent of $alpha$, which does not make sense physically.







integration complex-analysis






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asked Dec 10 '18 at 20:06









EwoudEwoud

22517




22517












  • $begingroup$
    You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
    $endgroup$
    – Zachary
    Dec 10 '18 at 20:25










  • $begingroup$
    I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
    $endgroup$
    – projectilemotion
    Dec 10 '18 at 20:57




















  • $begingroup$
    You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
    $endgroup$
    – Zachary
    Dec 10 '18 at 20:25










  • $begingroup$
    I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
    $endgroup$
    – projectilemotion
    Dec 10 '18 at 20:57


















$begingroup$
You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
$endgroup$
– Zachary
Dec 10 '18 at 20:25




$begingroup$
You could try differentiating $I$ with respect to $alpha$ twice and see where that takes you.
$endgroup$
– Zachary
Dec 10 '18 at 20:25












$begingroup$
I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
$endgroup$
– projectilemotion
Dec 10 '18 at 20:57






$begingroup$
I got a result using modified Struve functions and Modified Bessel functions: $$-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+frac{pi}{R}$$ I will write an answer, shortly.
$endgroup$
– projectilemotion
Dec 10 '18 at 20:57












3 Answers
3






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7












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In this answer, we use the following well-known general results:



$$K_{n}(z)=frac{sqrt{pi}}{Gamma(n+frac{1}{2})}(z/2)^nint_1^{infty} frac{e^{-zx}}{(x^2-1)^{1/2-n}}~dx tag{1}$$
And:
$$int K_0(z)~dz=frac{pi z}{2}(K_0(z)L_{-1}(z)+K_1(z)L_0(z))+C tag{2}$$
Where $K_n(z)$ is the modified Bessel function of the second kind and $L_n(z)$ is the modified Struve function. These results have been found here on equation (7) and here respectively.




We use differentiation under the integral sign:
$$I(alpha)=int_{-infty}^{infty} frac{1}{R^2+z^2}e^{-alphasqrt{R^2+z^2}}~dz$$
$$I'(alpha)=-int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz$$
The second integral can be computed using the substitution $t=sqrt{R^2+z^2}$ and the substitution $u=t/R$:
$$begin{align}int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz&=2int_{0}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz\&=2int_R^{infty} frac{e^{-at}}{sqrt{t^2-R^2}}~dt\&=2int_1^{infty} frac{e^{-alpha Ru}}{sqrt{u^2-1}}~du end{align}$$
Using equation $(1)$ with $n=0$ gives us a very simple result:
$$I'(alpha)=-2K_0(alpha R)$$
We can now integrate with respect to $alpha$, using equation $(2)$. We obtain:
$$I(alpha)=-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+C$$
Using that $I(0)=int_{-infty}^{infty} frac{1}{R^2+z^2}~dz=pi/R$, we obtain that $C=pi/R$. Thus, the result is:
$$I=frac{pi}{R}-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))$$



I highly doubt that one can simplify this further.






share|cite|improve this answer











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    2












    $begingroup$


    1. Differentiate w.r.t $alpha$

    2. Sub $z=R sinh(x)$

    3. do the integral over x $int_{mathbb{R}}dxexp(-ccosh(x))=K_0(c)$, $K_0(c)$ is a modified Bessel-function

    4. integrate back w.r.t. $alpha$
      Use $tfrac{2}{pi z}int dz K_0(z)= L_0(z)
      K_0(z)+L_{-1}(z)K_1(z)+C$
      , $L_b(z)$ is a so called Struve
      function

    5. Find Integration constant $C$ by noting that the inital integral is clearly $0$ for $alpha=infty$


    PS: Your contour Approach fails since you didn't take the branchcut $[i,iinfty]$ into account :/






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      When you integrate using residue theorem, you often integrate over a semicircumference with radius that goes to infinity. Remember that you have to integrate over ALL the semicircumference. Often the integral on the semicircumference (curved part) is 0, in this case it probably is not. You have to sum both integrals (real line AND semicircumference) to use residue theorem. Actually you can use other curves, like rectangles, the important thing is that you have to consider all the pieces of the curve.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
        $endgroup$
        – Ewoud
        Dec 10 '18 at 20:33












      • $begingroup$
        Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
        $endgroup$
        – tommy1996q
        Dec 10 '18 at 20:36













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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$


      In this answer, we use the following well-known general results:



      $$K_{n}(z)=frac{sqrt{pi}}{Gamma(n+frac{1}{2})}(z/2)^nint_1^{infty} frac{e^{-zx}}{(x^2-1)^{1/2-n}}~dx tag{1}$$
      And:
      $$int K_0(z)~dz=frac{pi z}{2}(K_0(z)L_{-1}(z)+K_1(z)L_0(z))+C tag{2}$$
      Where $K_n(z)$ is the modified Bessel function of the second kind and $L_n(z)$ is the modified Struve function. These results have been found here on equation (7) and here respectively.




      We use differentiation under the integral sign:
      $$I(alpha)=int_{-infty}^{infty} frac{1}{R^2+z^2}e^{-alphasqrt{R^2+z^2}}~dz$$
      $$I'(alpha)=-int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz$$
      The second integral can be computed using the substitution $t=sqrt{R^2+z^2}$ and the substitution $u=t/R$:
      $$begin{align}int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz&=2int_{0}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz\&=2int_R^{infty} frac{e^{-at}}{sqrt{t^2-R^2}}~dt\&=2int_1^{infty} frac{e^{-alpha Ru}}{sqrt{u^2-1}}~du end{align}$$
      Using equation $(1)$ with $n=0$ gives us a very simple result:
      $$I'(alpha)=-2K_0(alpha R)$$
      We can now integrate with respect to $alpha$, using equation $(2)$. We obtain:
      $$I(alpha)=-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+C$$
      Using that $I(0)=int_{-infty}^{infty} frac{1}{R^2+z^2}~dz=pi/R$, we obtain that $C=pi/R$. Thus, the result is:
      $$I=frac{pi}{R}-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))$$



      I highly doubt that one can simplify this further.






      share|cite|improve this answer











      $endgroup$


















        7












        $begingroup$


        In this answer, we use the following well-known general results:



        $$K_{n}(z)=frac{sqrt{pi}}{Gamma(n+frac{1}{2})}(z/2)^nint_1^{infty} frac{e^{-zx}}{(x^2-1)^{1/2-n}}~dx tag{1}$$
        And:
        $$int K_0(z)~dz=frac{pi z}{2}(K_0(z)L_{-1}(z)+K_1(z)L_0(z))+C tag{2}$$
        Where $K_n(z)$ is the modified Bessel function of the second kind and $L_n(z)$ is the modified Struve function. These results have been found here on equation (7) and here respectively.




        We use differentiation under the integral sign:
        $$I(alpha)=int_{-infty}^{infty} frac{1}{R^2+z^2}e^{-alphasqrt{R^2+z^2}}~dz$$
        $$I'(alpha)=-int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz$$
        The second integral can be computed using the substitution $t=sqrt{R^2+z^2}$ and the substitution $u=t/R$:
        $$begin{align}int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz&=2int_{0}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz\&=2int_R^{infty} frac{e^{-at}}{sqrt{t^2-R^2}}~dt\&=2int_1^{infty} frac{e^{-alpha Ru}}{sqrt{u^2-1}}~du end{align}$$
        Using equation $(1)$ with $n=0$ gives us a very simple result:
        $$I'(alpha)=-2K_0(alpha R)$$
        We can now integrate with respect to $alpha$, using equation $(2)$. We obtain:
        $$I(alpha)=-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+C$$
        Using that $I(0)=int_{-infty}^{infty} frac{1}{R^2+z^2}~dz=pi/R$, we obtain that $C=pi/R$. Thus, the result is:
        $$I=frac{pi}{R}-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))$$



        I highly doubt that one can simplify this further.






        share|cite|improve this answer











        $endgroup$
















          7












          7








          7





          $begingroup$


          In this answer, we use the following well-known general results:



          $$K_{n}(z)=frac{sqrt{pi}}{Gamma(n+frac{1}{2})}(z/2)^nint_1^{infty} frac{e^{-zx}}{(x^2-1)^{1/2-n}}~dx tag{1}$$
          And:
          $$int K_0(z)~dz=frac{pi z}{2}(K_0(z)L_{-1}(z)+K_1(z)L_0(z))+C tag{2}$$
          Where $K_n(z)$ is the modified Bessel function of the second kind and $L_n(z)$ is the modified Struve function. These results have been found here on equation (7) and here respectively.




          We use differentiation under the integral sign:
          $$I(alpha)=int_{-infty}^{infty} frac{1}{R^2+z^2}e^{-alphasqrt{R^2+z^2}}~dz$$
          $$I'(alpha)=-int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz$$
          The second integral can be computed using the substitution $t=sqrt{R^2+z^2}$ and the substitution $u=t/R$:
          $$begin{align}int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz&=2int_{0}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz\&=2int_R^{infty} frac{e^{-at}}{sqrt{t^2-R^2}}~dt\&=2int_1^{infty} frac{e^{-alpha Ru}}{sqrt{u^2-1}}~du end{align}$$
          Using equation $(1)$ with $n=0$ gives us a very simple result:
          $$I'(alpha)=-2K_0(alpha R)$$
          We can now integrate with respect to $alpha$, using equation $(2)$. We obtain:
          $$I(alpha)=-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+C$$
          Using that $I(0)=int_{-infty}^{infty} frac{1}{R^2+z^2}~dz=pi/R$, we obtain that $C=pi/R$. Thus, the result is:
          $$I=frac{pi}{R}-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))$$



          I highly doubt that one can simplify this further.






          share|cite|improve this answer











          $endgroup$




          In this answer, we use the following well-known general results:



          $$K_{n}(z)=frac{sqrt{pi}}{Gamma(n+frac{1}{2})}(z/2)^nint_1^{infty} frac{e^{-zx}}{(x^2-1)^{1/2-n}}~dx tag{1}$$
          And:
          $$int K_0(z)~dz=frac{pi z}{2}(K_0(z)L_{-1}(z)+K_1(z)L_0(z))+C tag{2}$$
          Where $K_n(z)$ is the modified Bessel function of the second kind and $L_n(z)$ is the modified Struve function. These results have been found here on equation (7) and here respectively.




          We use differentiation under the integral sign:
          $$I(alpha)=int_{-infty}^{infty} frac{1}{R^2+z^2}e^{-alphasqrt{R^2+z^2}}~dz$$
          $$I'(alpha)=-int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz$$
          The second integral can be computed using the substitution $t=sqrt{R^2+z^2}$ and the substitution $u=t/R$:
          $$begin{align}int_{-infty}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz&=2int_{0}^{infty} frac{e^{-alphasqrt{R^2+z^2}}}{sqrt{R^2+z^2}}~dz\&=2int_R^{infty} frac{e^{-at}}{sqrt{t^2-R^2}}~dt\&=2int_1^{infty} frac{e^{-alpha Ru}}{sqrt{u^2-1}}~du end{align}$$
          Using equation $(1)$ with $n=0$ gives us a very simple result:
          $$I'(alpha)=-2K_0(alpha R)$$
          We can now integrate with respect to $alpha$, using equation $(2)$. We obtain:
          $$I(alpha)=-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))+C$$
          Using that $I(0)=int_{-infty}^{infty} frac{1}{R^2+z^2}~dz=pi/R$, we obtain that $C=pi/R$. Thus, the result is:
          $$I=frac{pi}{R}-pialpha(L_{-1}(alpha R)K_0(alpha R)+L_0(alpha R)K_1(alpha R))$$



          I highly doubt that one can simplify this further.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 21:57

























          answered Dec 10 '18 at 21:09









          projectilemotionprojectilemotion

          11.4k62141




          11.4k62141























              2












              $begingroup$


              1. Differentiate w.r.t $alpha$

              2. Sub $z=R sinh(x)$

              3. do the integral over x $int_{mathbb{R}}dxexp(-ccosh(x))=K_0(c)$, $K_0(c)$ is a modified Bessel-function

              4. integrate back w.r.t. $alpha$
                Use $tfrac{2}{pi z}int dz K_0(z)= L_0(z)
                K_0(z)+L_{-1}(z)K_1(z)+C$
                , $L_b(z)$ is a so called Struve
                function

              5. Find Integration constant $C$ by noting that the inital integral is clearly $0$ for $alpha=infty$


              PS: Your contour Approach fails since you didn't take the branchcut $[i,iinfty]$ into account :/






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$


                1. Differentiate w.r.t $alpha$

                2. Sub $z=R sinh(x)$

                3. do the integral over x $int_{mathbb{R}}dxexp(-ccosh(x))=K_0(c)$, $K_0(c)$ is a modified Bessel-function

                4. integrate back w.r.t. $alpha$
                  Use $tfrac{2}{pi z}int dz K_0(z)= L_0(z)
                  K_0(z)+L_{-1}(z)K_1(z)+C$
                  , $L_b(z)$ is a so called Struve
                  function

                5. Find Integration constant $C$ by noting that the inital integral is clearly $0$ for $alpha=infty$


                PS: Your contour Approach fails since you didn't take the branchcut $[i,iinfty]$ into account :/






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$


                  1. Differentiate w.r.t $alpha$

                  2. Sub $z=R sinh(x)$

                  3. do the integral over x $int_{mathbb{R}}dxexp(-ccosh(x))=K_0(c)$, $K_0(c)$ is a modified Bessel-function

                  4. integrate back w.r.t. $alpha$
                    Use $tfrac{2}{pi z}int dz K_0(z)= L_0(z)
                    K_0(z)+L_{-1}(z)K_1(z)+C$
                    , $L_b(z)$ is a so called Struve
                    function

                  5. Find Integration constant $C$ by noting that the inital integral is clearly $0$ for $alpha=infty$


                  PS: Your contour Approach fails since you didn't take the branchcut $[i,iinfty]$ into account :/






                  share|cite|improve this answer









                  $endgroup$




                  1. Differentiate w.r.t $alpha$

                  2. Sub $z=R sinh(x)$

                  3. do the integral over x $int_{mathbb{R}}dxexp(-ccosh(x))=K_0(c)$, $K_0(c)$ is a modified Bessel-function

                  4. integrate back w.r.t. $alpha$
                    Use $tfrac{2}{pi z}int dz K_0(z)= L_0(z)
                    K_0(z)+L_{-1}(z)K_1(z)+C$
                    , $L_b(z)$ is a so called Struve
                    function

                  5. Find Integration constant $C$ by noting that the inital integral is clearly $0$ for $alpha=infty$


                  PS: Your contour Approach fails since you didn't take the branchcut $[i,iinfty]$ into account :/







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 21:08









                  BesselsbabeBesselsbabe

                  211




                  211























                      0












                      $begingroup$

                      When you integrate using residue theorem, you often integrate over a semicircumference with radius that goes to infinity. Remember that you have to integrate over ALL the semicircumference. Often the integral on the semicircumference (curved part) is 0, in this case it probably is not. You have to sum both integrals (real line AND semicircumference) to use residue theorem. Actually you can use other curves, like rectangles, the important thing is that you have to consider all the pieces of the curve.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
                        $endgroup$
                        – Ewoud
                        Dec 10 '18 at 20:33












                      • $begingroup$
                        Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
                        $endgroup$
                        – tommy1996q
                        Dec 10 '18 at 20:36


















                      0












                      $begingroup$

                      When you integrate using residue theorem, you often integrate over a semicircumference with radius that goes to infinity. Remember that you have to integrate over ALL the semicircumference. Often the integral on the semicircumference (curved part) is 0, in this case it probably is not. You have to sum both integrals (real line AND semicircumference) to use residue theorem. Actually you can use other curves, like rectangles, the important thing is that you have to consider all the pieces of the curve.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
                        $endgroup$
                        – Ewoud
                        Dec 10 '18 at 20:33












                      • $begingroup$
                        Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
                        $endgroup$
                        – tommy1996q
                        Dec 10 '18 at 20:36
















                      0












                      0








                      0





                      $begingroup$

                      When you integrate using residue theorem, you often integrate over a semicircumference with radius that goes to infinity. Remember that you have to integrate over ALL the semicircumference. Often the integral on the semicircumference (curved part) is 0, in this case it probably is not. You have to sum both integrals (real line AND semicircumference) to use residue theorem. Actually you can use other curves, like rectangles, the important thing is that you have to consider all the pieces of the curve.






                      share|cite|improve this answer











                      $endgroup$



                      When you integrate using residue theorem, you often integrate over a semicircumference with radius that goes to infinity. Remember that you have to integrate over ALL the semicircumference. Often the integral on the semicircumference (curved part) is 0, in this case it probably is not. You have to sum both integrals (real line AND semicircumference) to use residue theorem. Actually you can use other curves, like rectangles, the important thing is that you have to consider all the pieces of the curve.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 10 '18 at 20:24

























                      answered Dec 10 '18 at 20:18









                      tommy1996qtommy1996q

                      591415




                      591415












                      • $begingroup$
                        The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
                        $endgroup$
                        – Ewoud
                        Dec 10 '18 at 20:33












                      • $begingroup$
                        Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
                        $endgroup$
                        – tommy1996q
                        Dec 10 '18 at 20:36




















                      • $begingroup$
                        The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
                        $endgroup$
                        – Ewoud
                        Dec 10 '18 at 20:33












                      • $begingroup$
                        Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
                        $endgroup$
                        – tommy1996q
                        Dec 10 '18 at 20:36


















                      $begingroup$
                      The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
                      $endgroup$
                      – Ewoud
                      Dec 10 '18 at 20:33






                      $begingroup$
                      The semicircumference should go to 0 I thought, because my integrand has an upper bound by $1/z^2$, so it decays faster than than $1/z$, so this upper circumference integral should be zero.
                      $endgroup$
                      – Ewoud
                      Dec 10 '18 at 20:33














                      $begingroup$
                      Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
                      $endgroup$
                      – tommy1996q
                      Dec 10 '18 at 20:36






                      $begingroup$
                      Yeah I think I remember a result saying something like that. I’d suggest you to give it a try, though. Also, did you remember you had to multiply by the derivative of the parametrization?
                      $endgroup$
                      – tommy1996q
                      Dec 10 '18 at 20:36




















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