Solving $int^{+ infty}_{- infty} exp(-x-e^{-x})dx= frac{1}{c}$











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
    15 hours ago















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
    15 hours ago













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 15 hours ago









amWhy

191k27223437




191k27223437










asked 17 hours ago









Atstovas

293




293












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


















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
















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




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










1 Answer
1






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.






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














     

    draft saved


    draft discarded


















    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

























    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.






    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 17 hours ago









        Richard Martin

        1,2588




        1,2588






























             

            draft saved


            draft discarded



















































             


            draft saved


            draft discarded














            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





















































            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







            Popular posts from this blog

            'app-layout' is not a known element: how to share Component with different Modules

            Image
            0 My application was working fine until I decided to go with lazy loading: So, my shared component looks like this: import { Component, Renderer2 } from '@angular/core'; export interface FormModel { captcha?: string; } @Component({ selector: 'app-layout', templateUrl: './app-layout.component.html', styleUrls: ['./app-layout.css'] }) export class FullLayoutComponent { } app-layout.component.ts I am using this component in my Another component say school-home.component.html like this: <app-layout> </app-layout> and it was working fine until I created a Module school-home.module.ts like this: import { NgModule } from '@angular/core'; import { CommonModule } from '@angular/common'; import { SchoolhomeRoutingModule ...
            Read more

            android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d ...
            Read more

            WPF add header to Image with URL pettitions [duplicate]

            Image
            up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:...
            Read more