Change of variable induces divergence in the integral
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 17 hours ago
J.A
205
add a comment |
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 17 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
17 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 17 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 17 hours ago
J.A
205
asked 17 hours ago
J.A
205
edited 17 hours ago
asked 17 hours ago
J.A
205
asked 17 hours ago
J.A
205
asked 17 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
17 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
17 hours ago
1
1
You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
– user121049
17 hours ago
You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
– user121049
17 hours ago
add a comment |
1 Answer
1
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 17 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 17 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 17 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 17 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 17 hours ago
Richard Martin
1,2588
answered 17 hours ago
Richard Martin
1,2588
answered 17 hours ago
Richard Martin
1,2588
answered 17 hours ago
Richard Martin
1,2588
1,2588
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004705%2fchange-of-variable-induces-divergence-in-the-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
– user121049
17 hours ago