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 Nov 19 at 9:39
J.A
1375
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 Nov 19 at 9:39
J.A
1375
1
You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
– user121049
Nov 19 at 9:44
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 Nov 19 at 9:39
J.A
1375
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 Nov 19 at 9:39
J.A
1375
asked Nov 19 at 9:39
J.A
1375
edited Nov 19 at 9:51
asked Nov 19 at 9:39
J.A
1375
asked Nov 19 at 9:39
J.A
1375
asked Nov 19 at 9:39
J.A
1375
1375
1
You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
– user121049
Nov 19 at 9:44
add a comment |
1
You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
– user121049
Nov 19 at 9:44
1
1
You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
– user121049
Nov 19 at 9:44
You forgot to change your integration limits for starters. Plot $z(x)$ to see what is going on.
– user121049
Nov 19 at 9:44
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 Nov 19 at 9:45
Richard Martin
1,6328
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 Nov 19 at 9:45
Richard Martin
1,6328
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 Nov 19 at 9:45
Richard Martin
1,6328
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 Nov 19 at 9:45
Richard Martin
1,6328
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 Nov 19 at 9:45
Richard Martin
1,6328
answered Nov 19 at 9:45
Richard Martin
1,6328
answered Nov 19 at 9:45
Richard Martin
1,6328
answered Nov 19 at 9:45
Richard Martin
1,6328
1,6328
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Nov 19 at 9:44