Change of variable induces divergence in the integral











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I have an integral, with $a,b >0$ :



$int_0^infty e^{-a^2 x^2 (1-x/b)^2}dx$



It's not diverging. Now let's change the variable : $zrightarrow a x (1-x/b)$. The integral becomes :



$int_{-infty}^{z(ba/2)} e^{-z^2}frac{b}{sqrt{a} sqrt{a b^2 - 4 z}}dz$



which has a singularity in $z=ab^2/4$. How did that happen ?






integration convergence change-of-variable







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    You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
    – user121049
    8 hours ago















up vote
0
down vote

favorite












I have an integral, with $a,b >0$ :



$int_0^infty e^{-a^2 x^2 (1-x/b)^2}dx$



It's not diverging. Now let's change the variable : $zrightarrow a x (1-x/b)$. The integral becomes :



$int_{-infty}^{z(ba/2)} e^{-z^2}frac{b}{sqrt{a} sqrt{a b^2 - 4 z}}dz$



which has a singularity in $z=ab^2/4$. How did that happen ?






integration convergence change-of-variable







share|cite|improve this question




















  • 1




    You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
    – user121049
    8 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have an integral, with $a,b >0$ :



$int_0^infty e^{-a^2 x^2 (1-x/b)^2}dx$



It's not diverging. Now let's change the variable : $zrightarrow a x (1-x/b)$. The integral becomes :



$int_{-infty}^{z(ba/2)} e^{-z^2}frac{b}{sqrt{a} sqrt{a b^2 - 4 z}}dz$



which has a singularity in $z=ab^2/4$. How did that happen ?






integration convergence change-of-variable







share|cite|improve this question















I have an integral, with $a,b >0$ :



$int_0^infty e^{-a^2 x^2 (1-x/b)^2}dx$



It's not diverging. Now let's change the variable : $zrightarrow a x (1-x/b)$. The integral becomes :



$int_{-infty}^{z(ba/2)} e^{-z^2}frac{b}{sqrt{a} sqrt{a b^2 - 4 z}}dz$



which has a singularity in $z=ab^2/4$. How did that happen ?






integration convergence change-of-variable




integration convergence change-of-variable






share|cite|improve this question















share|cite|improve this question













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edited 8 hours ago

























asked 9 hours ago









J.A

205




205








  • 1




    You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
    – user121049
    8 hours ago














  • 1




    You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
    – user121049
    8 hours ago








1




1




You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
– user121049
8 hours ago




You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
– user121049
8 hours ago










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Your change of variable is not one-to-one, which invalidates the RHS. But, the existence of a singularity is not precluded by a change of variable, e.g. $$int_0^infty e^{-x^2/2} dx = int_0^infty e^{-y} (2y)^{-1/2} dy$$
because the integrand is still (Riemann) integrable.






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    up vote
    2
    down vote



    accepted










    Your change of variable is not one-to-one, which invalidates the RHS. But, the existence of a singularity is not precluded by a change of variable, e.g. $$int_0^infty e^{-x^2/2} dx = int_0^infty e^{-y} (2y)^{-1/2} dy$$
    because the integrand is still (Riemann) integrable.






    share|cite|improve this answer

























      up vote
      2
      down vote



      accepted










      Your change of variable is not one-to-one, which invalidates the RHS. But, the existence of a singularity is not precluded by a change of variable, e.g. $$int_0^infty e^{-x^2/2} dx = int_0^infty e^{-y} (2y)^{-1/2} dy$$
      because the integrand is still (Riemann) integrable.






      share|cite|improve this answer























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Your change of variable is not one-to-one, which invalidates the RHS. But, the existence of a singularity is not precluded by a change of variable, e.g. $$int_0^infty e^{-x^2/2} dx = int_0^infty e^{-y} (2y)^{-1/2} dy$$
        because the integrand is still (Riemann) integrable.






        share|cite|improve this answer












        Your change of variable is not one-to-one, which invalidates the RHS. But, the existence of a singularity is not precluded by a change of variable, e.g. $$int_0^infty e^{-x^2/2} dx = int_0^infty e^{-y} (2y)^{-1/2} dy$$
        because the integrand is still (Riemann) integrable.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 8 hours ago









        Richard Martin

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