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

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



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




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  • 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
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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























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        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
                        $$B=int_1^+inftyfrace^-tsqrtt,dtleqint_1^+inftye^-t , dt < infty$$






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                          $$int_0^+inftyfrace^-tsqrttdt=int_0^1frace^-tsqrtt , dt + int_1^+inftyfrace^-tsqrtt , dt=A+B$$
                          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

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                            204k23185460











                            answered 11 hours ago









                            user 108128

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