Determine whether the integral $ int^+infty_0frace^-t sqrt t , dt$ converges or diverges?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
4
down vote

favorite
2












$$ int^+infty_0frace^-t sqrt t , dt$$



I have computed it getting $$ frac2e3 therefore text it converges.$$ using integration by parts letting $$u = frac 1 sqrt t $$ and $$ dV= e^-t , dt$$
Is it the right way to do it ?




calculus improper-integrals




share|cite|improve this question





















  • Do you know of the gamma function?
    – TheSimpliFire
    11 hours ago






  • 1




    @TheSimpliFire no I don't
    – Mayar
    11 hours ago














up vote
4
down vote

favorite
2












$$ int^+infty_0frace^-t sqrt t , dt$$



I have computed it getting $$ frac2e3 therefore text it converges.$$ using integration by parts letting $$u = frac 1 sqrt t $$ and $$ dV= e^-t , dt$$
Is it the right way to do it ?




calculus improper-integrals




share|cite|improve this question





















  • Do you know of the gamma function?
    – TheSimpliFire
    11 hours ago






  • 1




    @TheSimpliFire no I don't
    – Mayar
    11 hours ago












up vote
4
down vote

favorite
2









up vote
4
down vote

favorite
2






2





$$ int^+infty_0frace^-t sqrt t , dt$$



I have computed it getting $$ frac2e3 therefore text it converges.$$ using integration by parts letting $$u = frac 1 sqrt t $$ and $$ dV= e^-t , dt$$
Is it the right way to do it ?




calculus improper-integrals




share|cite|improve this question













$$ int^+infty_0frace^-t sqrt t , dt$$



I have computed it getting $$ frac2e3 therefore text it converges.$$ using integration by parts letting $$u = frac 1 sqrt t $$ and $$ dV= e^-t , dt$$
Is it the right way to do it ?





calculus improper-integrals





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Asaf Karagila

291k31400731




291k31400731









asked 11 hours ago









Mayar

285




285











  • Do you know of the gamma function?
    – TheSimpliFire
    11 hours ago






  • 1




    @TheSimpliFire no I don't
    – Mayar
    11 hours ago
















  • Do you know of the gamma function?
    – TheSimpliFire
    11 hours ago






  • 1




    @TheSimpliFire no I don't
    – Mayar
    11 hours ago















Do you know of the gamma function?
– TheSimpliFire
11 hours ago




Do you know of the gamma function?
– TheSimpliFire
11 hours ago




1




1




@TheSimpliFire no I don't
– Mayar
11 hours ago




@TheSimpliFire no I don't
– Mayar
11 hours ago










4 Answers
4






active

oldest

votes

















up vote
10
down vote













If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero,
$$
int_0^1 frace^-tsqrt t,dtleq int_0^1frac1sqrt t,dt,
$$
which is convergent.



At infinity,
$$
int_1^infty frace^-tsqrt t,dtleqint_1^infty e^-t,dt<infty.
$$
So the integral converges.






share|cite|improve this answer




























    up vote
    6
    down vote













    $$I=int_0^inftyfrace^-tsqrttdt$$
    $u=sqrtt,,dt=2sqrttdu$
    $$I=2int_0^inftye^-u^2du=sqrtpi$$ As it is a standard integral



    EDIT:
    $$I=2int_0^inftye^-x^2dx$$
    then
    $$I^2=4left(int_0^inftye^-x^2dxright)^2=4left(int_0^inftye^-x^2right)left(int_0^inftye^-y^2dyright)=4int_0^inftyint_0^inftye^-(x^2+y^2)dxdy$$
    now we can use polar coordinates to simplify this.
    $x^2+y^2=r^2,$ and $dV=dxdy=rdrdtheta$ so our integral becomes:
    $$I^2=4int_0^fracpi2int_0^inftye^-r^2rdrdtheta$$
    now $u=-r^2$ so $fracdudr=-2r, therefore,dr=fracdu-2r$ and the integral becomes:
    $$I^2=-2int_0^fracpi2int_0^-inftye^ududtheta=2int_0^fracpi2int_-infty^0e^ududtheta=2int_0^fracpi2left[e^uright]_-infty^0dtheta=2int_0^fracpi2dtheta=2*fracpi2=pi$$
    so if $I^2=pi$ then $I=sqrtpi$






    share|cite|improve this answer























    • This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
      – Henry Lee
      11 hours ago










    • Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
      – Mayar
      6 hours ago











    • I don't really understand your notation, but I will add to my answer to show why this is the case
      – Henry Lee
      6 hours ago










    • Okay thank you .
      – Mayar
      6 hours ago











    • I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
      – Henry Lee
      6 hours ago


















    up vote
    3
    down vote













    As Henry Lee's answer mentions, you get $sqrtpi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+infty,$ then it converges. If it's not everywhere positive, and the meaning of convergence may become problematic. For example,
    $$
    lim_a,to,+inftyint_0^a fracsin x x , dx = frac pi 2
    $$
    and that is a sort of convergence, but
    $$
    intlimits_,x ,:, (sin x)/x , ge,0 fracsin x x,dx = +infty quadtext and quad intlimits_,x ,:, (sin x)/x , < ,0 fracsin x x,dx = -infty
    $$
    and some questions arise about when one should consider the thing convergent.



    But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus
    $$
    0 < int_0^1 frace^-tsqrt t , dt le int_0^1 frac 1 sqrt t , dt < +infty
    $$
    and
    $$
    0 < int_1^infty frace^-tsqrt t,dt le int_1^infty e^-t , dx = e^-1 < +infty
    $$
    so what you have converges.






    share|cite|improve this answer




























      up vote
      3
      down vote













      $$int_0^+inftyfrace^-tsqrttdt=int_0^1frace^-tsqrtt , dt + int_1^+inftyfrace^-tsqrtt , dt=A+B$$
      for $A$ when $e^-tsim1$ then
      $$A=int_0^1 frace^-tsqrtt,dtsimint_0^1 frac1sqrtt , dt < infty$$
      for $B$ when $tgeq1$ then
      $$B=int_1^+inftyfrace^-tsqrtt,dtleqint_1^+inftye^-t , dt < infty$$






      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: false,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        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%2f2873036%2fdetermine-whether-the-integral-int-infty-0-frace-t-sqrt-t-dt%23new-answer', 'question_page');

        );

        Post as a guest

















        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        10
        down vote













        If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero,
        $$
        int_0^1 frace^-tsqrt t,dtleq int_0^1frac1sqrt t,dt,
        $$
        which is convergent.



        At infinity,
        $$
        int_1^infty frace^-tsqrt t,dtleqint_1^infty e^-t,dt<infty.
        $$
        So the integral converges.






        share|cite|improve this answer

























          up vote
          10
          down vote













          If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero,
          $$
          int_0^1 frace^-tsqrt t,dtleq int_0^1frac1sqrt t,dt,
          $$
          which is convergent.



          At infinity,
          $$
          int_1^infty frace^-tsqrt t,dtleqint_1^infty e^-t,dt<infty.
          $$
          So the integral converges.






          share|cite|improve this answer























            up vote
            10
            down vote










            up vote
            10
            down vote









            If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero,
            $$
            int_0^1 frace^-tsqrt t,dtleq int_0^1frac1sqrt t,dt,
            $$
            which is convergent.



            At infinity,
            $$
            int_1^infty frace^-tsqrt t,dtleqint_1^infty e^-t,dt<infty.
            $$
            So the integral converges.






            share|cite|improve this answer













            If you simply want to decide convergence and you don't notice the trick in Henry Lee's answer, you can look at the following. Near zero,
            $$
            int_0^1 frace^-tsqrt t,dtleq int_0^1frac1sqrt t,dt,
            $$
            which is convergent.



            At infinity,
            $$
            int_1^infty frace^-tsqrt t,dtleqint_1^infty e^-t,dt<infty.
            $$
            So the integral converges.







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered 10 hours ago









            Martin Argerami

            115k1070164




            115k1070164




















                up vote
                6
                down vote













                $$I=int_0^inftyfrace^-tsqrttdt$$
                $u=sqrtt,,dt=2sqrttdu$
                $$I=2int_0^inftye^-u^2du=sqrtpi$$ As it is a standard integral



                EDIT:
                $$I=2int_0^inftye^-x^2dx$$
                then
                $$I^2=4left(int_0^inftye^-x^2dxright)^2=4left(int_0^inftye^-x^2right)left(int_0^inftye^-y^2dyright)=4int_0^inftyint_0^inftye^-(x^2+y^2)dxdy$$
                now we can use polar coordinates to simplify this.
                $x^2+y^2=r^2,$ and $dV=dxdy=rdrdtheta$ so our integral becomes:
                $$I^2=4int_0^fracpi2int_0^inftye^-r^2rdrdtheta$$
                now $u=-r^2$ so $fracdudr=-2r, therefore,dr=fracdu-2r$ and the integral becomes:
                $$I^2=-2int_0^fracpi2int_0^-inftye^ududtheta=2int_0^fracpi2int_-infty^0e^ududtheta=2int_0^fracpi2left[e^uright]_-infty^0dtheta=2int_0^fracpi2dtheta=2*fracpi2=pi$$
                so if $I^2=pi$ then $I=sqrtpi$






                share|cite|improve this answer























                • This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
                  – Henry Lee
                  11 hours ago










                • Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
                  – Mayar
                  6 hours ago











                • I don't really understand your notation, but I will add to my answer to show why this is the case
                  – Henry Lee
                  6 hours ago










                • Okay thank you .
                  – Mayar
                  6 hours ago











                • I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
                  – Henry Lee
                  6 hours ago















                up vote
                6
                down vote













                $$I=int_0^inftyfrace^-tsqrttdt$$
                $u=sqrtt,,dt=2sqrttdu$
                $$I=2int_0^inftye^-u^2du=sqrtpi$$ As it is a standard integral



                EDIT:
                $$I=2int_0^inftye^-x^2dx$$
                then
                $$I^2=4left(int_0^inftye^-x^2dxright)^2=4left(int_0^inftye^-x^2right)left(int_0^inftye^-y^2dyright)=4int_0^inftyint_0^inftye^-(x^2+y^2)dxdy$$
                now we can use polar coordinates to simplify this.
                $x^2+y^2=r^2,$ and $dV=dxdy=rdrdtheta$ so our integral becomes:
                $$I^2=4int_0^fracpi2int_0^inftye^-r^2rdrdtheta$$
                now $u=-r^2$ so $fracdudr=-2r, therefore,dr=fracdu-2r$ and the integral becomes:
                $$I^2=-2int_0^fracpi2int_0^-inftye^ududtheta=2int_0^fracpi2int_-infty^0e^ududtheta=2int_0^fracpi2left[e^uright]_-infty^0dtheta=2int_0^fracpi2dtheta=2*fracpi2=pi$$
                so if $I^2=pi$ then $I=sqrtpi$






                share|cite|improve this answer























                • This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
                  – Henry Lee
                  11 hours ago










                • Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
                  – Mayar
                  6 hours ago











                • I don't really understand your notation, but I will add to my answer to show why this is the case
                  – Henry Lee
                  6 hours ago










                • Okay thank you .
                  – Mayar
                  6 hours ago











                • I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
                  – Henry Lee
                  6 hours ago













                up vote
                6
                down vote










                up vote
                6
                down vote









                $$I=int_0^inftyfrace^-tsqrttdt$$
                $u=sqrtt,,dt=2sqrttdu$
                $$I=2int_0^inftye^-u^2du=sqrtpi$$ As it is a standard integral



                EDIT:
                $$I=2int_0^inftye^-x^2dx$$
                then
                $$I^2=4left(int_0^inftye^-x^2dxright)^2=4left(int_0^inftye^-x^2right)left(int_0^inftye^-y^2dyright)=4int_0^inftyint_0^inftye^-(x^2+y^2)dxdy$$
                now we can use polar coordinates to simplify this.
                $x^2+y^2=r^2,$ and $dV=dxdy=rdrdtheta$ so our integral becomes:
                $$I^2=4int_0^fracpi2int_0^inftye^-r^2rdrdtheta$$
                now $u=-r^2$ so $fracdudr=-2r, therefore,dr=fracdu-2r$ and the integral becomes:
                $$I^2=-2int_0^fracpi2int_0^-inftye^ududtheta=2int_0^fracpi2int_-infty^0e^ududtheta=2int_0^fracpi2left[e^uright]_-infty^0dtheta=2int_0^fracpi2dtheta=2*fracpi2=pi$$
                so if $I^2=pi$ then $I=sqrtpi$






                share|cite|improve this answer















                $$I=int_0^inftyfrace^-tsqrttdt$$
                $u=sqrtt,,dt=2sqrttdu$
                $$I=2int_0^inftye^-u^2du=sqrtpi$$ As it is a standard integral



                EDIT:
                $$I=2int_0^inftye^-x^2dx$$
                then
                $$I^2=4left(int_0^inftye^-x^2dxright)^2=4left(int_0^inftye^-x^2right)left(int_0^inftye^-y^2dyright)=4int_0^inftyint_0^inftye^-(x^2+y^2)dxdy$$
                now we can use polar coordinates to simplify this.
                $x^2+y^2=r^2,$ and $dV=dxdy=rdrdtheta$ so our integral becomes:
                $$I^2=4int_0^fracpi2int_0^inftye^-r^2rdrdtheta$$
                now $u=-r^2$ so $fracdudr=-2r, therefore,dr=fracdu-2r$ and the integral becomes:
                $$I^2=-2int_0^fracpi2int_0^-inftye^ududtheta=2int_0^fracpi2int_-infty^0e^ududtheta=2int_0^fracpi2left[e^uright]_-infty^0dtheta=2int_0^fracpi2dtheta=2*fracpi2=pi$$
                so if $I^2=pi$ then $I=sqrtpi$







                share|cite|improve this answer















                share|cite|improve this answer



                share|cite|improve this answer








                edited 6 hours ago


























                answered 11 hours ago









                Henry Lee

                47210




                47210











                • This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
                  – Henry Lee
                  11 hours ago










                • Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
                  – Mayar
                  6 hours ago











                • I don't really understand your notation, but I will add to my answer to show why this is the case
                  – Henry Lee
                  6 hours ago










                • Okay thank you .
                  – Mayar
                  6 hours ago











                • I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
                  – Henry Lee
                  6 hours ago

















                • This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
                  – Henry Lee
                  11 hours ago










                • Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
                  – Mayar
                  6 hours ago











                • I don't really understand your notation, but I will add to my answer to show why this is the case
                  – Henry Lee
                  6 hours ago










                • Okay thank you .
                  – Mayar
                  6 hours ago











                • I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
                  – Henry Lee
                  6 hours ago
















                This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
                – Henry Lee
                11 hours ago




                This can also be done using the gamma function then linking to factorial but this is a well known result so thought this way is easier
                – Henry Lee
                11 hours ago












                Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
                – Mayar
                6 hours ago





                Yes this is easier , could you show me how you computed $$ sqrt( pi)$$ though , I know that by getting$$ int v dU$$ its gonna be $$ = 1/2 .(I) $$
                – Mayar
                6 hours ago













                I don't really understand your notation, but I will add to my answer to show why this is the case
                – Henry Lee
                6 hours ago




                I don't really understand your notation, but I will add to my answer to show why this is the case
                – Henry Lee
                6 hours ago












                Okay thank you .
                – Mayar
                6 hours ago





                Okay thank you .
                – Mayar
                6 hours ago













                I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
                – Henry Lee
                6 hours ago





                I have now added a proof. This uses polar coordinates which you may not be familiar with, and is commonly known as the Gaussian integral
                – Henry Lee
                6 hours ago











                up vote
                3
                down vote













                As Henry Lee's answer mentions, you get $sqrtpi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+infty,$ then it converges. If it's not everywhere positive, and the meaning of convergence may become problematic. For example,
                $$
                lim_a,to,+inftyint_0^a fracsin x x , dx = frac pi 2
                $$
                and that is a sort of convergence, but
                $$
                intlimits_,x ,:, (sin x)/x , ge,0 fracsin x x,dx = +infty quadtext and quad intlimits_,x ,:, (sin x)/x , < ,0 fracsin x x,dx = -infty
                $$
                and some questions arise about when one should consider the thing convergent.



                But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus
                $$
                0 < int_0^1 frace^-tsqrt t , dt le int_0^1 frac 1 sqrt t , dt < +infty
                $$
                and
                $$
                0 < int_1^infty frace^-tsqrt t,dt le int_1^infty e^-t , dx = e^-1 < +infty
                $$
                so what you have converges.






                share|cite|improve this answer

























                  up vote
                  3
                  down vote













                  As Henry Lee's answer mentions, you get $sqrtpi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+infty,$ then it converges. If it's not everywhere positive, and the meaning of convergence may become problematic. For example,
                  $$
                  lim_a,to,+inftyint_0^a fracsin x x , dx = frac pi 2
                  $$
                  and that is a sort of convergence, but
                  $$
                  intlimits_,x ,:, (sin x)/x , ge,0 fracsin x x,dx = +infty quadtext and quad intlimits_,x ,:, (sin x)/x , < ,0 fracsin x x,dx = -infty
                  $$
                  and some questions arise about when one should consider the thing convergent.



                  But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus
                  $$
                  0 < int_0^1 frace^-tsqrt t , dt le int_0^1 frac 1 sqrt t , dt < +infty
                  $$
                  and
                  $$
                  0 < int_1^infty frace^-tsqrt t,dt le int_1^infty e^-t , dx = e^-1 < +infty
                  $$
                  so what you have converges.






                  share|cite|improve this answer























                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    As Henry Lee's answer mentions, you get $sqrtpi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+infty,$ then it converges. If it's not everywhere positive, and the meaning of convergence may become problematic. For example,
                    $$
                    lim_a,to,+inftyint_0^a fracsin x x , dx = frac pi 2
                    $$
                    and that is a sort of convergence, but
                    $$
                    intlimits_,x ,:, (sin x)/x , ge,0 fracsin x x,dx = +infty quadtext and quad intlimits_,x ,:, (sin x)/x , < ,0 fracsin x x,dx = -infty
                    $$
                    and some questions arise about when one should consider the thing convergent.



                    But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus
                    $$
                    0 < int_0^1 frace^-tsqrt t , dt le int_0^1 frac 1 sqrt t , dt < +infty
                    $$
                    and
                    $$
                    0 < int_1^infty frace^-tsqrt t,dt le int_1^infty e^-t , dx = e^-1 < +infty
                    $$
                    so what you have converges.






                    share|cite|improve this answer













                    As Henry Lee's answer mentions, you get $sqrtpi,$ so $2e/3$ is wrong. But if the function is everywhere positive and you computed it correctly and get a finite number rather than $+infty,$ then it converges. If it's not everywhere positive, and the meaning of convergence may become problematic. For example,
                    $$
                    lim_a,to,+inftyint_0^a fracsin x x , dx = frac pi 2
                    $$
                    and that is a sort of convergence, but
                    $$
                    intlimits_,x ,:, (sin x)/x , ge,0 fracsin x x,dx = +infty quadtext and quad intlimits_,x ,:, (sin x)/x , < ,0 fracsin x x,dx = -infty
                    $$
                    and some questions arise about when one should consider the thing convergent.



                    But one thing to bear in mind is that the question of whether something converges or not is often simpler than the question of what it converges to. Thus
                    $$
                    0 < int_0^1 frace^-tsqrt t , dt le int_0^1 frac 1 sqrt t , dt < +infty
                    $$
                    and
                    $$
                    0 < int_1^infty frace^-tsqrt t,dt le int_1^infty e^-t , dx = e^-1 < +infty
                    $$
                    so what you have converges.







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered 10 hours ago









                    Michael Hardy

                    204k23185460




                    204k23185460




















                        up vote
                        3
                        down vote













                        $$int_0^+inftyfrace^-tsqrttdt=int_0^1frace^-tsqrtt , dt + int_1^+inftyfrace^-tsqrtt , dt=A+B$$
                        for $A$ when $e^-tsim1$ then
                        $$A=int_0^1 frace^-tsqrtt,dtsimint_0^1 frac1sqrtt , dt < infty$$
                        for $B$ when $tgeq1$ then
                        $$B=int_1^+inftyfrace^-tsqrtt,dtleqint_1^+inftye^-t , dt < infty$$






                        share|cite|improve this answer



























                          up vote
                          3
                          down vote













                          $$int_0^+inftyfrace^-tsqrttdt=int_0^1frace^-tsqrtt , dt + int_1^+inftyfrace^-tsqrtt , dt=A+B$$
                          for $A$ when $e^-tsim1$ then
                          $$A=int_0^1 frace^-tsqrtt,dtsimint_0^1 frac1sqrtt , dt < infty$$
                          for $B$ when $tgeq1$ then
                          $$B=int_1^+inftyfrace^-tsqrtt,dtleqint_1^+inftye^-t , dt < infty$$






                          share|cite|improve this answer

























                            up vote
                            3
                            down vote










                            up vote
                            3
                            down vote









                            $$int_0^+inftyfrace^-tsqrttdt=int_0^1frace^-tsqrtt , dt + int_1^+inftyfrace^-tsqrtt , dt=A+B$$
                            for $A$ when $e^-tsim1$ then
                            $$A=int_0^1 frace^-tsqrtt,dtsimint_0^1 frac1sqrtt , dt < infty$$
                            for $B$ when $tgeq1$ then
                            $$B=int_1^+inftyfrace^-tsqrtt,dtleqint_1^+inftye^-t , dt < infty$$






                            share|cite|improve this answer















                            $$int_0^+inftyfrace^-tsqrttdt=int_0^1frace^-tsqrtt , dt + int_1^+inftyfrace^-tsqrtt , dt=A+B$$
                            for $A$ when $e^-tsim1$ then
                            $$A=int_0^1 frace^-tsqrtt,dtsimint_0^1 frac1sqrtt , dt < infty$$
                            for $B$ when $tgeq1$ then
                            $$B=int_1^+inftyfrace^-tsqrtt,dtleqint_1^+inftye^-t , dt < infty$$







                            share|cite|improve this answer















                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 10 hours ago









                            Michael Hardy

                            204k23185460




                            204k23185460











                            answered 11 hours ago









                            user 108128

                            18.6k41544




                            18.6k41544






















                                 

                                draft saved


                                draft discarded


























                                 


                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function ()
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873036%2fdetermine-whether-the-integral-int-infty-0-frace-t-sqrt-t-dt%23new-answer', 'question_page');

                                );

                                Post as a guest
































































                                Comments

                                Post a Comment

                                Popular posts from this blog

                                What is the equation of a 3D cone with generalised tilt?

                                Image
                                Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an
                                Read more

                                Color the edges and diagonals of a regular polygon

                                Image
                                Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne
                                Read more

                                Relationship between determinant of matrix and determinant of adjoint?

                                Image
                                Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/… – Hector Blandin Nov 17 &
                                Read more