Solving $int^{+ infty}_{- infty} exp(-x-e^{-x})dx= frac{1}{c}$
up vote
1
down vote
favorite
I need to solve an equation.
$int^{+ infty}_{- infty} exp(-x-e^{-x})dx= frac{1}{c}$
I know that
$int^{+infty}_{-infty} exp(-x-e^{-x})dx= lim_{a rightarrow -infty} int^{0}_{a} exp(-x-e^{-x})dx + lim_{b {rightarrow infty}} int^{b}_{0} exp(-x-e^{-x})dx$
Can somebody explain step by step how to integrate $int lim_{a rightarrow -infty} int^{a}_{0} exp(-x-e^{-x})dx $
I used variable exchange method and I got
$lim_{a rightarrow -infty} (e-e^{-a})$.
Is it right? But how to solve this limit?
probability integration limits
edited Nov 19 at 11:47
amWhy
191k27223439
asked Nov 19 at 10:06
Atstovas
466
add a comment |
up vote
1
down vote
favorite
I need to solve an equation.
$int^{+ infty}_{- infty} exp(-x-e^{-x})dx= frac{1}{c}$
I know that
$int^{+infty}_{-infty} exp(-x-e^{-x})dx= lim_{a rightarrow -infty} int^{0}_{a} exp(-x-e^{-x})dx + lim_{b {rightarrow infty}} int^{b}_{0} exp(-x-e^{-x})dx$
Can somebody explain step by step how to integrate $int lim_{a rightarrow -infty} int^{a}_{0} exp(-x-e^{-x})dx $
I used variable exchange method and I got
$lim_{a rightarrow -infty} (e-e^{-a})$.
Is it right? But how to solve this limit?
probability integration limits
edited Nov 19 at 11:47
amWhy
191k27223439
asked Nov 19 at 10:06
Atstovas
466
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
Nov 19 at 11:30
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need to solve an equation.
$int^{+ infty}_{- infty} exp(-x-e^{-x})dx= frac{1}{c}$
I know that
$int^{+infty}_{-infty} exp(-x-e^{-x})dx= lim_{a rightarrow -infty} int^{0}_{a} exp(-x-e^{-x})dx + lim_{b {rightarrow infty}} int^{b}_{0} exp(-x-e^{-x})dx$
Can somebody explain step by step how to integrate $int lim_{a rightarrow -infty} int^{a}_{0} exp(-x-e^{-x})dx $
I used variable exchange method and I got
$lim_{a rightarrow -infty} (e-e^{-a})$.
Is it right? But how to solve this limit?
probability integration limits
edited Nov 19 at 11:47
amWhy
191k27223439
asked Nov 19 at 10:06
Atstovas
466
I need to solve an equation.
$int^{+ infty}_{- infty} exp(-x-e^{-x})dx= frac{1}{c}$
I know that
$int^{+infty}_{-infty} exp(-x-e^{-x})dx= lim_{a rightarrow -infty} int^{0}_{a} exp(-x-e^{-x})dx + lim_{b {rightarrow infty}} int^{b}_{0} exp(-x-e^{-x})dx$
Can somebody explain step by step how to integrate $int lim_{a rightarrow -infty} int^{a}_{0} exp(-x-e^{-x})dx $
I used variable exchange method and I got
$lim_{a rightarrow -infty} (e-e^{-a})$.
Is it right? But how to solve this limit?
probability integration limits
probability integration limits
edited Nov 19 at 11:47
amWhy
191k27223439
asked Nov 19 at 10:06
Atstovas
466
edited Nov 19 at 11:47
amWhy
191k27223439
asked Nov 19 at 10:06
Atstovas
466
edited Nov 19 at 11:47
amWhy
191k27223439
edited Nov 19 at 11:47
amWhy
191k27223439
edited Nov 19 at 11:47
amWhy
191k27223439
191k27223439
asked Nov 19 at 10:06
Atstovas
466
asked Nov 19 at 10:06
Atstovas
466
asked Nov 19 at 10:06
Atstovas
466
466
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
Nov 19 at 11:30
add a comment |
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
Nov 19 at 11:30
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
Nov 19 at 11:30
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
Nov 19 at 11:30
add a comment |
2 Answers
2
active
oldest
votes
up vote
6
down vote
This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.
answered Nov 19 at 10:12
Richard Martin
1,6328
add a comment |
up vote
1
down vote
we have:
$$I=intexpleft(-x-e^{-x}right)dx$$
$$=int e^{-x-e^{-x}}dx$$
$u=e^{-x}$ so $dx=frac{du}{-e^{-x}}$
$$I=-int e^{-u}du=e^{-u}+C=e^{-e^{-x}}+C$$
now put in limits
edited Nov 21 at 0:16
Batominovski
32.1k23190
answered Nov 21 at 0:09
Henry Lee
1,692218
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.
answered Nov 19 at 10:12
Richard Martin
1,6328
add a comment |
up vote
6
down vote
This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.
answered Nov 19 at 10:12
Richard Martin
1,6328
add a comment |
up vote
6
down vote
up vote
6
down vote
This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.
answered Nov 19 at 10:12
Richard Martin
1,6328
This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.
answered Nov 19 at 10:12
Richard Martin
1,6328
answered Nov 19 at 10:12
Richard Martin
1,6328
answered Nov 19 at 10:12
Richard Martin
1,6328
answered Nov 19 at 10:12
Richard Martin
1,6328
1,6328
add a comment |
add a comment |
up vote
1
down vote
we have:
$$I=intexpleft(-x-e^{-x}right)dx$$
$$=int e^{-x-e^{-x}}dx$$
$u=e^{-x}$ so $dx=frac{du}{-e^{-x}}$
$$I=-int e^{-u}du=e^{-u}+C=e^{-e^{-x}}+C$$
now put in limits
edited Nov 21 at 0:16
Batominovski
32.1k23190
answered Nov 21 at 0:09
Henry Lee
1,692218
add a comment |
up vote
1
down vote
we have:
$$I=intexpleft(-x-e^{-x}right)dx$$
$$=int e^{-x-e^{-x}}dx$$
$u=e^{-x}$ so $dx=frac{du}{-e^{-x}}$
$$I=-int e^{-u}du=e^{-u}+C=e^{-e^{-x}}+C$$
now put in limits
edited Nov 21 at 0:16
Batominovski
32.1k23190
answered Nov 21 at 0:09
Henry Lee
1,692218
add a comment |
up vote
1
down vote
up vote
1
down vote
we have:
$$I=intexpleft(-x-e^{-x}right)dx$$
$$=int e^{-x-e^{-x}}dx$$
$u=e^{-x}$ so $dx=frac{du}{-e^{-x}}$
$$I=-int e^{-u}du=e^{-u}+C=e^{-e^{-x}}+C$$
now put in limits
edited Nov 21 at 0:16
Batominovski
32.1k23190
answered Nov 21 at 0:09
Henry Lee
1,692218
we have:
$$I=intexpleft(-x-e^{-x}right)dx$$
$$=int e^{-x-e^{-x}}dx$$
$u=e^{-x}$ so $dx=frac{du}{-e^{-x}}$
$$I=-int e^{-u}du=e^{-u}+C=e^{-e^{-x}}+C$$
now put in limits
edited Nov 21 at 0:16
Batominovski
32.1k23190
answered Nov 21 at 0:09
Henry Lee
1,692218
edited Nov 21 at 0:16
Batominovski
32.1k23190
edited Nov 21 at 0:16
Batominovski
32.1k23190
edited Nov 21 at 0:16
Batominovski
32.1k23190
32.1k23190
answered Nov 21 at 0:09
Henry Lee
1,692218
answered Nov 21 at 0:09
Henry Lee
1,692218
answered Nov 21 at 0:09
Henry Lee
1,692218
1,692218
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%2f3004740%2fsolving-int-infty-infty-exp-x-e-xdx-frac1c%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
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
Nov 19 at 11:30