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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 yesterday
amWhy
191k27223437
asked yesterday
Atstovas
283
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 yesterday
amWhy
191k27223437
asked yesterday
Atstovas
283
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
yesterday
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 yesterday
amWhy
191k27223437
asked yesterday
Atstovas
283
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 yesterday
amWhy
191k27223437
asked yesterday
Atstovas
283
edited yesterday
amWhy
191k27223437
asked yesterday
Atstovas
283
edited yesterday
amWhy
191k27223437
edited yesterday
amWhy
191k27223437
edited yesterday
amWhy
191k27223437
191k27223437
asked yesterday
Atstovas
283
asked yesterday
Atstovas
283
asked yesterday
Atstovas
283
283
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
yesterday
add a comment |
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
yesterday
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
yesterday
Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
– Easymode44
yesterday
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 yesterday
Richard Martin
1,3238
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 yesterday
Richard Martin
1,3238
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 yesterday
Richard Martin
1,3238
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 yesterday
Richard Martin
1,3238
This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
answered yesterday
Richard Martin
1,3238
1,3238
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
yesterday