Solving $int^{+ infty}_{- infty} exp(-x-e^{-x})dx= frac{1}{c}$
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1
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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
191k28223439
asked Nov 19 at 10:06
Atstovas
587
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
191k28223439
asked Nov 19 at 10:06
Atstovas
587
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
191k28223439
asked Nov 19 at 10:06
Atstovas
587
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
191k28223439
asked Nov 19 at 10:06
Atstovas
587
edited Nov 19 at 11:47
amWhy
191k28223439
asked Nov 19 at 10:06
Atstovas
587
edited Nov 19 at 11:47
amWhy
191k28223439
edited Nov 19 at 11:47
amWhy
191k28223439
edited Nov 19 at 11:47
amWhy
191k28223439
191k28223439
asked Nov 19 at 10:06
Atstovas
587
asked Nov 19 at 10:06
Atstovas
587
asked Nov 19 at 10:06
Atstovas
587
587
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
1,63918
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
32.9k23192
answered Nov 21 at 0:09
Henry Lee
1,682218
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
1,63918
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
1,63918
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
1,63918
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
1,63918
answered Nov 19 at 10:12
Richard Martin
1,63918
answered Nov 19 at 10:12
Richard Martin
1,63918
answered Nov 19 at 10:12
Richard Martin
1,63918
1,63918
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
32.9k23192
answered Nov 21 at 0:09
Henry Lee
1,682218
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
32.9k23192
answered Nov 21 at 0:09
Henry Lee
1,682218
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
32.9k23192
answered Nov 21 at 0:09
Henry Lee
1,682218
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
32.9k23192
answered Nov 21 at 0:09
Henry Lee
1,682218
edited Nov 21 at 0:16
Batominovski
32.9k23192
edited Nov 21 at 0:16
Batominovski
32.9k23192
edited Nov 21 at 0:16
Batominovski
32.9k23192
32.9k23192
answered Nov 21 at 0:09
Henry Lee
1,682218
answered Nov 21 at 0:09
Henry Lee
1,682218
answered Nov 21 at 0:09
Henry Lee
1,682218
1,682218
add a comment |
add a comment |
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Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
Nov 19 at 11:30