Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.












3












$begingroup$



Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.




I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.



Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.



Could someone check if I'm good till now and maybe help me continue? Thanks!






integration probability-theory random-variables conditional-expectation expected-value







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are good :-) no mistake.
    $endgroup$
    – Math-fun
    Jan 8 at 20:05










  • $begingroup$
    You received two answers with different numerical results. Have you checked out the discrepancy?
    $endgroup$
    – herb steinberg
    Jan 10 at 18:28






  • 1




    $begingroup$
    @rapidracim I have corrected my answer and there is no discrepancy between the two results.
    $endgroup$
    – herb steinberg
    Jan 14 at 20:46










  • $begingroup$
    @herbsteinberg I checked your edit, I get it now.
    $endgroup$
    – rapidracim
    Jan 14 at 21:39
















3












$begingroup$



Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.




I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.



Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.



Could someone check if I'm good till now and maybe help me continue? Thanks!






integration probability-theory random-variables conditional-expectation expected-value







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You are good :-) no mistake.
    $endgroup$
    – Math-fun
    Jan 8 at 20:05










  • $begingroup$
    You received two answers with different numerical results. Have you checked out the discrepancy?
    $endgroup$
    – herb steinberg
    Jan 10 at 18:28






  • 1




    $begingroup$
    @rapidracim I have corrected my answer and there is no discrepancy between the two results.
    $endgroup$
    – herb steinberg
    Jan 14 at 20:46










  • $begingroup$
    @herbsteinberg I checked your edit, I get it now.
    $endgroup$
    – rapidracim
    Jan 14 at 21:39














3












3








3





$begingroup$



Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.




I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.



Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.



Could someone check if I'm good till now and maybe help me continue? Thanks!






integration probability-theory random-variables conditional-expectation expected-value







share|cite|improve this question











$endgroup$





Compute $mathbb{E}big[exp(XY+X)big]$ where $X, Y$ are independent uniformly distributed over $[0,1]$ r.v's.




I first computed $mathbb{E}big[exp(XY+X)mid Xbig]$.



Because $X$ and $Y$ are independent then $$mathbb{E}big[exp(XY+X)mid Xbig] = phi(X),,$$
where $$phi(x) =mathbb{E}big[exp(xY+x)big] = int_0^1 exp(xy+x)dy = frac{exp x(exp x - 1)}{x}.$$
so $$mathbb{E}big[exp(XY+X)big] = int_0^1 frac{exp x(exp x - 1)}{x} dx,,$$
which I don't know how to compute.



Could someone check if I'm good till now and maybe help me continue? Thanks!






integration probability-theory random-variables conditional-expectation expected-value




integration probability-theory random-variables conditional-expectation expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 22:06









Batominovski

1




1










asked Jan 8 at 19:57









rapidracimrapidracim

1,6141319




1,6141319








  • 1




    $begingroup$
    You are good :-) no mistake.
    $endgroup$
    – Math-fun
    Jan 8 at 20:05










  • $begingroup$
    You received two answers with different numerical results. Have you checked out the discrepancy?
    $endgroup$
    – herb steinberg
    Jan 10 at 18:28






  • 1




    $begingroup$
    @rapidracim I have corrected my answer and there is no discrepancy between the two results.
    $endgroup$
    – herb steinberg
    Jan 14 at 20:46










  • $begingroup$
    @herbsteinberg I checked your edit, I get it now.
    $endgroup$
    – rapidracim
    Jan 14 at 21:39














  • 1




    $begingroup$
    You are good :-) no mistake.
    $endgroup$
    – Math-fun
    Jan 8 at 20:05










  • $begingroup$
    You received two answers with different numerical results. Have you checked out the discrepancy?
    $endgroup$
    – herb steinberg
    Jan 10 at 18:28






  • 1




    $begingroup$
    @rapidracim I have corrected my answer and there is no discrepancy between the two results.
    $endgroup$
    – herb steinberg
    Jan 14 at 20:46










  • $begingroup$
    @herbsteinberg I checked your edit, I get it now.
    $endgroup$
    – rapidracim
    Jan 14 at 21:39








1




1




$begingroup$
You are good :-) no mistake.
$endgroup$
– Math-fun
Jan 8 at 20:05




$begingroup$
You are good :-) no mistake.
$endgroup$
– Math-fun
Jan 8 at 20:05












$begingroup$
You received two answers with different numerical results. Have you checked out the discrepancy?
$endgroup$
– herb steinberg
Jan 10 at 18:28




$begingroup$
You received two answers with different numerical results. Have you checked out the discrepancy?
$endgroup$
– herb steinberg
Jan 10 at 18:28




1




1




$begingroup$
@rapidracim I have corrected my answer and there is no discrepancy between the two results.
$endgroup$
– herb steinberg
Jan 14 at 20:46




$begingroup$
@rapidracim I have corrected my answer and there is no discrepancy between the two results.
$endgroup$
– herb steinberg
Jan 14 at 20:46












$begingroup$
@herbsteinberg I checked your edit, I get it now.
$endgroup$
– rapidracim
Jan 14 at 21:39




$begingroup$
@herbsteinberg I checked your edit, I get it now.
$endgroup$
– rapidracim
Jan 14 at 21:39










2 Answers
2






active

oldest

votes


















2












$begingroup$

You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
$$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
(Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
In other words,
$$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
This also shows that
$$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$



The required integral is
begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
&=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
&=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
Now
$$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
This gives
$$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
    I calculated this result and got I=3.0591165.

    In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$

    Net result$=2.365969319$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @ Batominovski What was the edit? I didn't see any changes!
      $endgroup$
      – herb steinberg
      Jan 8 at 22:34











    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',
    autoActivateHeartbeat: false,
    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%2f3066648%2fcompute-mathbbe-big-expxyx-big-where-x-y-are-independent-uniformly%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
    $$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
    (Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
    In other words,
    $$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
    This also shows that
    $$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$



    The required integral is
    begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
    &=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
    &=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
    Now
    $$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
    This gives
    $$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
      $$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
      (Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
      In other words,
      $$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
      This also shows that
      $$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$



      The required integral is
      begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
      &=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
      &=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
      Now
      $$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
      This gives
      $$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
        $$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
        (Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
        In other words,
        $$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
        This also shows that
        $$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$



        The required integral is
        begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
        &=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
        &=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
        Now
        $$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
        This gives
        $$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$






        share|cite|improve this answer











        $endgroup$



        You have done great so far. But I dont think there is an elementary expression for the last integral. But you can note that the exponential integral $operatorname{Ei}$ is given by
        $$operatorname{Ei}(t)=mathrel{-!!!!!!;!!int}_{-infty}^{t}frac{e^{s}}{s}ds.$$
        (Here $mathrel{-!!!!!;!!int}$ is the Cauchy principal value integral.)
        In other words,
        $$intfrac{e^{s}}{s}ds=operatorname{Ei}(s)+C.$$
        This also shows that
        $$intfrac{e^{ks}}{s}ds=operatorname{Ei}(ks)+C.$$



        The required integral is
        begin{align}I&=int_0^1frac{e^x(e^x-1)}{x}dx=lim_{epsilonsearrow0}int_{epsilon}^1left(frac{e^{2x}}{x}dx-int_epsilon^1frac{e^x}{x}dxright)\
        &=lim_{epsilonsearrow0}Big(big(operatorname{Ei}(2)-operatorname{Ei}(2epsilon)big)-big(operatorname{Ei}(1)-operatorname{Ei}(epsilon)big)Big)\
        &=operatorname{Ei}(2)-operatorname{Ei}(1)-lim_{epsilonsearrow0}big(operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)big).end{align}
        Now
        $$operatorname{Ei}(2epsilon)-operatorname{Ei}(epsilon)=int_{epsilon}^{2epsilon}frac{e^s}{s}ds=int_epsilon^{2epsilon}frac{1+O(epsilon)}{s}ds=int_{epsilon}^{2epsilon}frac{ds}{s}+O(epsilon)=ln 2+O(epsilon).$$
        This gives
        $$I=operatorname{Ei}(2)-operatorname{Ei}(1)-ln2approx 2.366.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 8 at 20:58

























        answered Jan 8 at 20:50







        user614671






























            0












            $begingroup$

            You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
            I calculated this result and got I=3.0591165.

            In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$

            Net result$=2.365969319$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @ Batominovski What was the edit? I didn't see any changes!
              $endgroup$
              – herb steinberg
              Jan 8 at 22:34
















            0












            $begingroup$

            You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
            I calculated this result and got I=3.0591165.

            In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$

            Net result$=2.365969319$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @ Batominovski What was the edit? I didn't see any changes!
              $endgroup$
              – herb steinberg
              Jan 8 at 22:34














            0












            0








            0





            $begingroup$

            You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
            I calculated this result and got I=3.0591165.

            In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$

            Net result$=2.365969319$.






            share|cite|improve this answer











            $endgroup$



            You can't get an explicit answer. However let $u=exp(x)$ then $du=exp(x)dx$ and your integral becomes $int_1^e frac{u-1}{ln(u)}du=text{li}(e^2)-text{li}(e)=int_e^{e^2}frac{1}{ln(u)}du$. Unfortunately there is no explicit formula for $text{li}(u)$.
            I calculated this result and got I=3.0591165.

            In addition, due to the singularity of the integrand at $u=1$, it is necessary to subtract $lim_{epsilonto 0}int_{(1+epsilon)}^{(1+epsilon)^2}frac{1}{ln(u)}du=ln(2)=0.693147181$

            Net result$=2.365969319$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 14 at 20:44

























            answered Jan 8 at 21:06









            herb steinbergherb steinberg

            2,5932310




            2,5932310












            • $begingroup$
              @ Batominovski What was the edit? I didn't see any changes!
              $endgroup$
              – herb steinberg
              Jan 8 at 22:34


















            • $begingroup$
              @ Batominovski What was the edit? I didn't see any changes!
              $endgroup$
              – herb steinberg
              Jan 8 at 22:34
















            $begingroup$
            @ Batominovski What was the edit? I didn't see any changes!
            $endgroup$
            – herb steinberg
            Jan 8 at 22:34




            $begingroup$
            @ Batominovski What was the edit? I didn't see any changes!
            $endgroup$
            – herb steinberg
            Jan 8 at 22:34


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066648%2fcompute-mathbbe-big-expxyx-big-where-x-y-are-independent-uniformly%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