Solving $int^{+ infty}_{- infty} exp(-x-e^{-x})dx= frac{1}{c}$
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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 18 hours ago
amWhy
191k27223437
asked 20 hours ago
Atstovas
293
add a comment |
up vote
0
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 18 hours ago
amWhy
191k27223437
asked 20 hours ago
Atstovas
293
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
19 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
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 18 hours ago
amWhy
191k27223437
asked 20 hours ago
Atstovas
293
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 18 hours ago
amWhy
191k27223437
asked 20 hours ago
Atstovas
293
edited 18 hours ago
amWhy
191k27223437
asked 20 hours ago
Atstovas
293
edited 18 hours ago
amWhy
191k27223437
edited 18 hours ago
amWhy
191k27223437
edited 18 hours ago
amWhy
191k27223437
191k27223437
asked 20 hours ago
Atstovas
293
asked 20 hours ago
Atstovas
293
asked 20 hours ago
Atstovas
293
293
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
19 hours ago
add a comment |
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
19 hours ago
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
19 hours ago
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
19 hours ago
add a comment |
1 Answer
1
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oldest
votes
up vote
4
down vote
This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.
answered 20 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
4
down vote
This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.
answered 20 hours ago
Richard Martin
1,2588
add a comment |
up vote
4
down vote
This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.
answered 20 hours ago
Richard Martin
1,2588
add a comment |
up vote
4
down vote
up vote
4
down vote
This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.
answered 20 hours ago
Richard Martin
1,2588
This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.
answered 20 hours ago
Richard Martin
1,2588
answered 20 hours ago
Richard Martin
1,2588
answered 20 hours ago
Richard Martin
1,2588
answered 20 hours ago
Richard Martin
1,2588
1,2588
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
19 hours ago