Quick question about an article about Watson triple integral

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So, I've recently read about Watson's triple integral, in particular i was interested in the first one. All this comes from: http://mathworld.wolfram.com/WatsonsTripleIntegrals.html. Now I've got through the derivation for
$$I_1=int_0^piint_0^piint_0^pifracdudvdw1-cosucosvcosw$$ (I will be ignoring the $pi^3$ term). Referencing to the link, I'm quite unsure about the step from (23) to (24). We have that:
$$4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dphi1+r^4sin^4thetacos^2thetasin^2phicos^2phi=$$
$$=4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dphi1+frac14r^4sin^4thetacos^2thetasin^22phi=$$
The OP says: we write $psi=2phi$ and continues:
$$=4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dpsi1+frac14r^4sin^4thetacos^2thetasin^2psi$$
But shouldn't also $dpsi=2dphi$ and the bounds of the first integral change to $0,pi$? The result of this integral, given in the link, which is $frac14Gamma^4(frac14)$ should be correct according to more articles. What am i missing?




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    So, I've recently read about Watson's triple integral, in particular i was interested in the first one. All this comes from: http://mathworld.wolfram.com/WatsonsTripleIntegrals.html. Now I've got through the derivation for
    $$I_1=int_0^piint_0^piint_0^pifracdudvdw1-cosucosvcosw$$ (I will be ignoring the $pi^3$ term). Referencing to the link, I'm quite unsure about the step from (23) to (24). We have that:
    $$4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dphi1+r^4sin^4thetacos^2thetasin^2phicos^2phi=$$
    $$=4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dphi1+frac14r^4sin^4thetacos^2thetasin^22phi=$$
    The OP says: we write $psi=2phi$ and continues:
    $$=4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dpsi1+frac14r^4sin^4thetacos^2thetasin^2psi$$
    But shouldn't also $dpsi=2dphi$ and the bounds of the first integral change to $0,pi$? The result of this integral, given in the link, which is $frac14Gamma^4(frac14)$ should be correct according to more articles. What am i missing?




    definite-integrals multiple-integral




    share|cite|improve this question





















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      So, I've recently read about Watson's triple integral, in particular i was interested in the first one. All this comes from: http://mathworld.wolfram.com/WatsonsTripleIntegrals.html. Now I've got through the derivation for
      $$I_1=int_0^piint_0^piint_0^pifracdudvdw1-cosucosvcosw$$ (I will be ignoring the $pi^3$ term). Referencing to the link, I'm quite unsure about the step from (23) to (24). We have that:
      $$4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dphi1+r^4sin^4thetacos^2thetasin^2phicos^2phi=$$
      $$=4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dphi1+frac14r^4sin^4thetacos^2thetasin^22phi=$$
      The OP says: we write $psi=2phi$ and continues:
      $$=4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dpsi1+frac14r^4sin^4thetacos^2thetasin^2psi$$
      But shouldn't also $dpsi=2dphi$ and the bounds of the first integral change to $0,pi$? The result of this integral, given in the link, which is $frac14Gamma^4(frac14)$ should be correct according to more articles. What am i missing?




      definite-integrals multiple-integral




      share|cite|improve this question











      So, I've recently read about Watson's triple integral, in particular i was interested in the first one. All this comes from: http://mathworld.wolfram.com/WatsonsTripleIntegrals.html. Now I've got through the derivation for
      $$I_1=int_0^piint_0^piint_0^pifracdudvdw1-cosucosvcosw$$ (I will be ignoring the $pi^3$ term). Referencing to the link, I'm quite unsure about the step from (23) to (24). We have that:
      $$4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dphi1+r^4sin^4thetacos^2thetasin^2phicos^2phi=$$
      $$=4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dphi1+frac14r^4sin^4thetacos^2thetasin^22phi=$$
      The OP says: we write $psi=2phi$ and continues:
      $$=4int_0^pi/2int_0^pi/2int_0^inftyfracsinthetadrdtheta dpsi1+frac14r^4sin^4thetacos^2thetasin^2psi$$
      But shouldn't also $dpsi=2dphi$ and the bounds of the first integral change to $0,pi$? The result of this integral, given in the link, which is $frac14Gamma^4(frac14)$ should be correct according to more articles. What am i missing?





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