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







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  • Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
    – Easymode44
    yesterday















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







share|cite|improve this question
























  • Are you sure about your limit expansion? Should they not be $a rightarrow infty$ and $b rightarrow -infty$?
    – Easymode44
    yesterday













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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









amWhy

191k27223437




191k27223437










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


















  • 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










1 Answer
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This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.






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    1 Answer
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    1 Answer
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    up vote
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    This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.






    share|cite|improve this answer

























      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.






      share|cite|improve this answer























        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.






        share|cite|improve this answer












        This isn't an integral equation---it's just doing an integral. Change variable $u=e^{-x}$, and you get 1.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        Richard Martin

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