Contour for $int_0^infty arctan(z) e^{-z^2},dz$ or some variant












5














I'm trying to practice my contour integration skills and got interested in the following integral:
$$int_0^infty arctan(z) e^{-z^2},dz$$
I know that the usual way to calculate integrals on $[0,infty)$ is to use the keyhole-contour, but the problem in this case is that $arctan(z)$ has branch points at $z=pm i$, with branch cuts usually chosen on $[i,iinfty)$ and $[-i,-iinfty)$, which I think means that the keyhole contour doesn't work in this case. I have tried a rectangle contour made up of $C:[0,R]cup [R+i/2] cup [R+i/2,i/2] cup [i/2,0]$ but that didn't work out. Because my function is holomorphic on $mathbb{C}setminus [i,iinfty) cap [-i,-iinfty)$ there also wouldn't be any residues to calculate, which doesn't necessarily have to be a problem as Cauchy's theorem could be used to try and compute the integral.



If I'm not mistaken, I would then get somerhing like $int_0^infty f(x)-f(x+i/2),dx=0$. One difficulty I ran into is simplify $f(z+i/2)$ (where $f(z)=arctan(x) e^{-z^2}$) into some other form such as $alpha f(z)+beta g(z)$ for some $g(z)$ whose integral can be calculated on $[0,infty)$ and $alpha, betainmathbb{C}$. This would let me then solve for $int_0^infty f(x),dx$.



I also thought that, if that makes it easier, we could extend the range of integration to $mathbb{R}$, as long as we found an odd function $q(a,x)$, such that $q(0,x)=1$ and compute



$$lim_{ato 0} frac{1}{2} int_{-infty}^infty q(a, x) arctan(x)e^{-x^2},dx$$
We could either use Cauchy's/the Residue theorem depending on whether $q(z)$ has poles or not. I have not been able to use this approach.



Any ideas?










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  • The value of the integral is $$G_{3, 2}^{1, 3} left( 1 middle| {1, 1, frac 3 2 atop 1, frac 1 2} right),$$ $G$ being the Meijer G-function. It's unlikely to have any significantly simpler closed form, it doesn't even seem to be expressible as a sum of generalized hypergeometric functions.
    – Maxim
    Jan 3 at 1:46












  • Are we allowed to use any other method to solve this integral? 'cause the Cauchy theorem doesn't seem to be applicable here.........
    – Mostafa Ayaz
    Jan 4 at 15:13
















5














I'm trying to practice my contour integration skills and got interested in the following integral:
$$int_0^infty arctan(z) e^{-z^2},dz$$
I know that the usual way to calculate integrals on $[0,infty)$ is to use the keyhole-contour, but the problem in this case is that $arctan(z)$ has branch points at $z=pm i$, with branch cuts usually chosen on $[i,iinfty)$ and $[-i,-iinfty)$, which I think means that the keyhole contour doesn't work in this case. I have tried a rectangle contour made up of $C:[0,R]cup [R+i/2] cup [R+i/2,i/2] cup [i/2,0]$ but that didn't work out. Because my function is holomorphic on $mathbb{C}setminus [i,iinfty) cap [-i,-iinfty)$ there also wouldn't be any residues to calculate, which doesn't necessarily have to be a problem as Cauchy's theorem could be used to try and compute the integral.



If I'm not mistaken, I would then get somerhing like $int_0^infty f(x)-f(x+i/2),dx=0$. One difficulty I ran into is simplify $f(z+i/2)$ (where $f(z)=arctan(x) e^{-z^2}$) into some other form such as $alpha f(z)+beta g(z)$ for some $g(z)$ whose integral can be calculated on $[0,infty)$ and $alpha, betainmathbb{C}$. This would let me then solve for $int_0^infty f(x),dx$.



I also thought that, if that makes it easier, we could extend the range of integration to $mathbb{R}$, as long as we found an odd function $q(a,x)$, such that $q(0,x)=1$ and compute



$$lim_{ato 0} frac{1}{2} int_{-infty}^infty q(a, x) arctan(x)e^{-x^2},dx$$
We could either use Cauchy's/the Residue theorem depending on whether $q(z)$ has poles or not. I have not been able to use this approach.



Any ideas?










share|cite|improve this question
























  • The value of the integral is $$G_{3, 2}^{1, 3} left( 1 middle| {1, 1, frac 3 2 atop 1, frac 1 2} right),$$ $G$ being the Meijer G-function. It's unlikely to have any significantly simpler closed form, it doesn't even seem to be expressible as a sum of generalized hypergeometric functions.
    – Maxim
    Jan 3 at 1:46












  • Are we allowed to use any other method to solve this integral? 'cause the Cauchy theorem doesn't seem to be applicable here.........
    – Mostafa Ayaz
    Jan 4 at 15:13














5












5








5


2





I'm trying to practice my contour integration skills and got interested in the following integral:
$$int_0^infty arctan(z) e^{-z^2},dz$$
I know that the usual way to calculate integrals on $[0,infty)$ is to use the keyhole-contour, but the problem in this case is that $arctan(z)$ has branch points at $z=pm i$, with branch cuts usually chosen on $[i,iinfty)$ and $[-i,-iinfty)$, which I think means that the keyhole contour doesn't work in this case. I have tried a rectangle contour made up of $C:[0,R]cup [R+i/2] cup [R+i/2,i/2] cup [i/2,0]$ but that didn't work out. Because my function is holomorphic on $mathbb{C}setminus [i,iinfty) cap [-i,-iinfty)$ there also wouldn't be any residues to calculate, which doesn't necessarily have to be a problem as Cauchy's theorem could be used to try and compute the integral.



If I'm not mistaken, I would then get somerhing like $int_0^infty f(x)-f(x+i/2),dx=0$. One difficulty I ran into is simplify $f(z+i/2)$ (where $f(z)=arctan(x) e^{-z^2}$) into some other form such as $alpha f(z)+beta g(z)$ for some $g(z)$ whose integral can be calculated on $[0,infty)$ and $alpha, betainmathbb{C}$. This would let me then solve for $int_0^infty f(x),dx$.



I also thought that, if that makes it easier, we could extend the range of integration to $mathbb{R}$, as long as we found an odd function $q(a,x)$, such that $q(0,x)=1$ and compute



$$lim_{ato 0} frac{1}{2} int_{-infty}^infty q(a, x) arctan(x)e^{-x^2},dx$$
We could either use Cauchy's/the Residue theorem depending on whether $q(z)$ has poles or not. I have not been able to use this approach.



Any ideas?










share|cite|improve this question















I'm trying to practice my contour integration skills and got interested in the following integral:
$$int_0^infty arctan(z) e^{-z^2},dz$$
I know that the usual way to calculate integrals on $[0,infty)$ is to use the keyhole-contour, but the problem in this case is that $arctan(z)$ has branch points at $z=pm i$, with branch cuts usually chosen on $[i,iinfty)$ and $[-i,-iinfty)$, which I think means that the keyhole contour doesn't work in this case. I have tried a rectangle contour made up of $C:[0,R]cup [R+i/2] cup [R+i/2,i/2] cup [i/2,0]$ but that didn't work out. Because my function is holomorphic on $mathbb{C}setminus [i,iinfty) cap [-i,-iinfty)$ there also wouldn't be any residues to calculate, which doesn't necessarily have to be a problem as Cauchy's theorem could be used to try and compute the integral.



If I'm not mistaken, I would then get somerhing like $int_0^infty f(x)-f(x+i/2),dx=0$. One difficulty I ran into is simplify $f(z+i/2)$ (where $f(z)=arctan(x) e^{-z^2}$) into some other form such as $alpha f(z)+beta g(z)$ for some $g(z)$ whose integral can be calculated on $[0,infty)$ and $alpha, betainmathbb{C}$. This would let me then solve for $int_0^infty f(x),dx$.



I also thought that, if that makes it easier, we could extend the range of integration to $mathbb{R}$, as long as we found an odd function $q(a,x)$, such that $q(0,x)=1$ and compute



$$lim_{ato 0} frac{1}{2} int_{-infty}^infty q(a, x) arctan(x)e^{-x^2},dx$$
We could either use Cauchy's/the Residue theorem depending on whether $q(z)$ has poles or not. I have not been able to use this approach.



Any ideas?







integration complex-analysis definite-integrals contour-integration complex-integration






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share|cite|improve this question













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share|cite|improve this question








edited Dec 28 '18 at 6:48







Zachary

















asked Dec 28 '18 at 6:42









ZacharyZachary

2,3071214




2,3071214












  • The value of the integral is $$G_{3, 2}^{1, 3} left( 1 middle| {1, 1, frac 3 2 atop 1, frac 1 2} right),$$ $G$ being the Meijer G-function. It's unlikely to have any significantly simpler closed form, it doesn't even seem to be expressible as a sum of generalized hypergeometric functions.
    – Maxim
    Jan 3 at 1:46












  • Are we allowed to use any other method to solve this integral? 'cause the Cauchy theorem doesn't seem to be applicable here.........
    – Mostafa Ayaz
    Jan 4 at 15:13


















  • The value of the integral is $$G_{3, 2}^{1, 3} left( 1 middle| {1, 1, frac 3 2 atop 1, frac 1 2} right),$$ $G$ being the Meijer G-function. It's unlikely to have any significantly simpler closed form, it doesn't even seem to be expressible as a sum of generalized hypergeometric functions.
    – Maxim
    Jan 3 at 1:46












  • Are we allowed to use any other method to solve this integral? 'cause the Cauchy theorem doesn't seem to be applicable here.........
    – Mostafa Ayaz
    Jan 4 at 15:13
















The value of the integral is $$G_{3, 2}^{1, 3} left( 1 middle| {1, 1, frac 3 2 atop 1, frac 1 2} right),$$ $G$ being the Meijer G-function. It's unlikely to have any significantly simpler closed form, it doesn't even seem to be expressible as a sum of generalized hypergeometric functions.
– Maxim
Jan 3 at 1:46






The value of the integral is $$G_{3, 2}^{1, 3} left( 1 middle| {1, 1, frac 3 2 atop 1, frac 1 2} right),$$ $G$ being the Meijer G-function. It's unlikely to have any significantly simpler closed form, it doesn't even seem to be expressible as a sum of generalized hypergeometric functions.
– Maxim
Jan 3 at 1:46














Are we allowed to use any other method to solve this integral? 'cause the Cauchy theorem doesn't seem to be applicable here.........
– Mostafa Ayaz
Jan 4 at 15:13




Are we allowed to use any other method to solve this integral? 'cause the Cauchy theorem doesn't seem to be applicable here.........
– Mostafa Ayaz
Jan 4 at 15:13










1 Answer
1






active

oldest

votes


















1














Applying of the Residue theorem is a very hard task, because the function $e^{-x^2}$ is not bounded when $zto icdot infty.$



On the other hand, parametric method can be used.



Let us consider the integral
$$begin{align}
&I(p) = intlimits_0^inftyarctan pz, e^{-z^2},mathrm dz,\[4pt]
&I'(p) = intlimits_0^inftydfrac{ze^{-z^2}}{p^2z^2+1},mathrm dz
= dfrac12 p^{-2}intlimits_0^inftydfrac{e^{-z^2}}{z^2+p^{-2}},mathrm dz^2.\[4pt]
end{align}$$

Using known definite integral
$$int_0^inftydfrac{e^{-ay}}{b+y},mathrm dy=e^{ab}mathrm {E}_1(ab),$$
one can get
$$I'(p)=dfrac12 p^{-2}e^{p^{-2}}mathrm {E}_1(-p^2),$$
$$I(p) = dfrac12 intlimits_0^p p^{-2}e^{p^{-2}}mathrm {E}_1(-p^2) dp
= left|p=frac1t,, t=frac1p,,mathrm dt=-frac{mathrm dp}{p^2}right|
= dfrac12intlimits_{p^{-1}}^infty e^{t^2}mathrm {E}_1(t^2), dt,$$

$$boxed{I(1) = dfrac12intlimits_{1}^infty e^{t^2}mathrm {E}_1(t^2), dt}.$$
Numeric calculations give the same result $I(1)approx0.40978$ for the both expressions, but this result can not presented via elementary functions.






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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Applying of the Residue theorem is a very hard task, because the function $e^{-x^2}$ is not bounded when $zto icdot infty.$



    On the other hand, parametric method can be used.



    Let us consider the integral
    $$begin{align}
    &I(p) = intlimits_0^inftyarctan pz, e^{-z^2},mathrm dz,\[4pt]
    &I'(p) = intlimits_0^inftydfrac{ze^{-z^2}}{p^2z^2+1},mathrm dz
    = dfrac12 p^{-2}intlimits_0^inftydfrac{e^{-z^2}}{z^2+p^{-2}},mathrm dz^2.\[4pt]
    end{align}$$

    Using known definite integral
    $$int_0^inftydfrac{e^{-ay}}{b+y},mathrm dy=e^{ab}mathrm {E}_1(ab),$$
    one can get
    $$I'(p)=dfrac12 p^{-2}e^{p^{-2}}mathrm {E}_1(-p^2),$$
    $$I(p) = dfrac12 intlimits_0^p p^{-2}e^{p^{-2}}mathrm {E}_1(-p^2) dp
    = left|p=frac1t,, t=frac1p,,mathrm dt=-frac{mathrm dp}{p^2}right|
    = dfrac12intlimits_{p^{-1}}^infty e^{t^2}mathrm {E}_1(t^2), dt,$$

    $$boxed{I(1) = dfrac12intlimits_{1}^infty e^{t^2}mathrm {E}_1(t^2), dt}.$$
    Numeric calculations give the same result $I(1)approx0.40978$ for the both expressions, but this result can not presented via elementary functions.






    share|cite|improve this answer




























      1














      Applying of the Residue theorem is a very hard task, because the function $e^{-x^2}$ is not bounded when $zto icdot infty.$



      On the other hand, parametric method can be used.



      Let us consider the integral
      $$begin{align}
      &I(p) = intlimits_0^inftyarctan pz, e^{-z^2},mathrm dz,\[4pt]
      &I'(p) = intlimits_0^inftydfrac{ze^{-z^2}}{p^2z^2+1},mathrm dz
      = dfrac12 p^{-2}intlimits_0^inftydfrac{e^{-z^2}}{z^2+p^{-2}},mathrm dz^2.\[4pt]
      end{align}$$

      Using known definite integral
      $$int_0^inftydfrac{e^{-ay}}{b+y},mathrm dy=e^{ab}mathrm {E}_1(ab),$$
      one can get
      $$I'(p)=dfrac12 p^{-2}e^{p^{-2}}mathrm {E}_1(-p^2),$$
      $$I(p) = dfrac12 intlimits_0^p p^{-2}e^{p^{-2}}mathrm {E}_1(-p^2) dp
      = left|p=frac1t,, t=frac1p,,mathrm dt=-frac{mathrm dp}{p^2}right|
      = dfrac12intlimits_{p^{-1}}^infty e^{t^2}mathrm {E}_1(t^2), dt,$$

      $$boxed{I(1) = dfrac12intlimits_{1}^infty e^{t^2}mathrm {E}_1(t^2), dt}.$$
      Numeric calculations give the same result $I(1)approx0.40978$ for the both expressions, but this result can not presented via elementary functions.






      share|cite|improve this answer


























        1












        1








        1






        Applying of the Residue theorem is a very hard task, because the function $e^{-x^2}$ is not bounded when $zto icdot infty.$



        On the other hand, parametric method can be used.



        Let us consider the integral
        $$begin{align}
        &I(p) = intlimits_0^inftyarctan pz, e^{-z^2},mathrm dz,\[4pt]
        &I'(p) = intlimits_0^inftydfrac{ze^{-z^2}}{p^2z^2+1},mathrm dz
        = dfrac12 p^{-2}intlimits_0^inftydfrac{e^{-z^2}}{z^2+p^{-2}},mathrm dz^2.\[4pt]
        end{align}$$

        Using known definite integral
        $$int_0^inftydfrac{e^{-ay}}{b+y},mathrm dy=e^{ab}mathrm {E}_1(ab),$$
        one can get
        $$I'(p)=dfrac12 p^{-2}e^{p^{-2}}mathrm {E}_1(-p^2),$$
        $$I(p) = dfrac12 intlimits_0^p p^{-2}e^{p^{-2}}mathrm {E}_1(-p^2) dp
        = left|p=frac1t,, t=frac1p,,mathrm dt=-frac{mathrm dp}{p^2}right|
        = dfrac12intlimits_{p^{-1}}^infty e^{t^2}mathrm {E}_1(t^2), dt,$$

        $$boxed{I(1) = dfrac12intlimits_{1}^infty e^{t^2}mathrm {E}_1(t^2), dt}.$$
        Numeric calculations give the same result $I(1)approx0.40978$ for the both expressions, but this result can not presented via elementary functions.






        share|cite|improve this answer














        Applying of the Residue theorem is a very hard task, because the function $e^{-x^2}$ is not bounded when $zto icdot infty.$



        On the other hand, parametric method can be used.



        Let us consider the integral
        $$begin{align}
        &I(p) = intlimits_0^inftyarctan pz, e^{-z^2},mathrm dz,\[4pt]
        &I'(p) = intlimits_0^inftydfrac{ze^{-z^2}}{p^2z^2+1},mathrm dz
        = dfrac12 p^{-2}intlimits_0^inftydfrac{e^{-z^2}}{z^2+p^{-2}},mathrm dz^2.\[4pt]
        end{align}$$

        Using known definite integral
        $$int_0^inftydfrac{e^{-ay}}{b+y},mathrm dy=e^{ab}mathrm {E}_1(ab),$$
        one can get
        $$I'(p)=dfrac12 p^{-2}e^{p^{-2}}mathrm {E}_1(-p^2),$$
        $$I(p) = dfrac12 intlimits_0^p p^{-2}e^{p^{-2}}mathrm {E}_1(-p^2) dp
        = left|p=frac1t,, t=frac1p,,mathrm dt=-frac{mathrm dp}{p^2}right|
        = dfrac12intlimits_{p^{-1}}^infty e^{t^2}mathrm {E}_1(t^2), dt,$$

        $$boxed{I(1) = dfrac12intlimits_{1}^infty e^{t^2}mathrm {E}_1(t^2), dt}.$$
        Numeric calculations give the same result $I(1)approx0.40978$ for the both expressions, but this result can not presented via elementary functions.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 20:52

























        answered Jan 2 at 22:04









        Yuri NegometyanovYuri Negometyanov

        11k1728




        11k1728






























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