Mixing
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
asked 9 hours ago
J.A
205
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0
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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
asked 9 hours ago
J.A
205
1
You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
– user121049
8 hours ago
add a comment |
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
asked 9 hours ago
J.A
205
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
asked 9 hours ago
J.A
205
asked 9 hours ago
J.A
205
edited 8 hours ago
asked 9 hours ago
J.A
205
asked 9 hours ago
J.A
205
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
add a comment |
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
add a comment |
1 Answer
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2
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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.
answered 8 hours ago
Richard Martin
1,2588
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 8 hours ago
Richard Martin
1,2588
add a comment |
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.
answered 8 hours ago
Richard Martin
1,2588
add a comment |
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.
answered 8 hours ago
Richard Martin
1,2588
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.
answered 8 hours ago
Richard Martin
1,2588
answered 8 hours ago
Richard Martin
1,2588
answered 8 hours ago
Richard Martin
1,2588
answered 8 hours ago
Richard Martin
1,2588
1,2588
add a comment |
add a comment |
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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