definite integration - solution breakdown












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$begingroup$


$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$



Unfortunately i cannot follow the steps. I guess integration by parts has been used?



How can Gamma density approach be applied here?






integration probability-distributions definite-integrals distribution-theory gamma-function







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  • $begingroup$
    Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
    $endgroup$
    – Luke Mathwalker
    Dec 12 '18 at 18:35


















0












$begingroup$


$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$



Unfortunately i cannot follow the steps. I guess integration by parts has been used?



How can Gamma density approach be applied here?






integration probability-distributions definite-integrals distribution-theory gamma-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
    $endgroup$
    – Luke Mathwalker
    Dec 12 '18 at 18:35
















0












0








0


1



$begingroup$


$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$



Unfortunately i cannot follow the steps. I guess integration by parts has been used?



How can Gamma density approach be applied here?






integration probability-distributions definite-integrals distribution-theory gamma-function







share|cite|improve this question











$endgroup$




$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$



Unfortunately i cannot follow the steps. I guess integration by parts has been used?



How can Gamma density approach be applied here?






integration probability-distributions definite-integrals distribution-theory gamma-function




integration probability-distributions definite-integrals distribution-theory gamma-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 18:48







user1607

















asked Dec 12 '18 at 18:25









user1607user1607

1718




1718












  • $begingroup$
    Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
    $endgroup$
    – Luke Mathwalker
    Dec 12 '18 at 18:35




















  • $begingroup$
    Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
    $endgroup$
    – Luke Mathwalker
    Dec 12 '18 at 18:35


















$begingroup$
Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
$endgroup$
– Luke Mathwalker
Dec 12 '18 at 18:35






$begingroup$
Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
$endgroup$
– Luke Mathwalker
Dec 12 '18 at 18:35












1 Answer
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$begingroup$

by parts, with $a=y+1$



$u=x$ and $v'=e^{-ax}$.



$$int_0^Xxe^{-ax}dx=$$



$$Bigl[xfrac{-1}{a}e^{-ax}Bigr]_0^X-int_0^Xfrac{-1}{a}e^{-ax}dx$$



$$frac{Xe^{-aX}}{a}+frac{1}{a}Bigl[frac{-1}{a}e^{-ax}Bigr]_0^X=$$



$$frac{Xe^{-aX}}{a}+frac{1}{a^2}-frac{e^{-aX}}{a^2}$$



now take the limit when $Xto+infty$.






share|cite|improve this answer









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    $begingroup$

    by parts, with $a=y+1$



    $u=x$ and $v'=e^{-ax}$.



    $$int_0^Xxe^{-ax}dx=$$



    $$Bigl[xfrac{-1}{a}e^{-ax}Bigr]_0^X-int_0^Xfrac{-1}{a}e^{-ax}dx$$



    $$frac{Xe^{-aX}}{a}+frac{1}{a}Bigl[frac{-1}{a}e^{-ax}Bigr]_0^X=$$



    $$frac{Xe^{-aX}}{a}+frac{1}{a^2}-frac{e^{-aX}}{a^2}$$



    now take the limit when $Xto+infty$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      by parts, with $a=y+1$



      $u=x$ and $v'=e^{-ax}$.



      $$int_0^Xxe^{-ax}dx=$$



      $$Bigl[xfrac{-1}{a}e^{-ax}Bigr]_0^X-int_0^Xfrac{-1}{a}e^{-ax}dx$$



      $$frac{Xe^{-aX}}{a}+frac{1}{a}Bigl[frac{-1}{a}e^{-ax}Bigr]_0^X=$$



      $$frac{Xe^{-aX}}{a}+frac{1}{a^2}-frac{e^{-aX}}{a^2}$$



      now take the limit when $Xto+infty$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        by parts, with $a=y+1$



        $u=x$ and $v'=e^{-ax}$.



        $$int_0^Xxe^{-ax}dx=$$



        $$Bigl[xfrac{-1}{a}e^{-ax}Bigr]_0^X-int_0^Xfrac{-1}{a}e^{-ax}dx$$



        $$frac{Xe^{-aX}}{a}+frac{1}{a}Bigl[frac{-1}{a}e^{-ax}Bigr]_0^X=$$



        $$frac{Xe^{-aX}}{a}+frac{1}{a^2}-frac{e^{-aX}}{a^2}$$



        now take the limit when $Xto+infty$.






        share|cite|improve this answer









        $endgroup$



        by parts, with $a=y+1$



        $u=x$ and $v'=e^{-ax}$.



        $$int_0^Xxe^{-ax}dx=$$



        $$Bigl[xfrac{-1}{a}e^{-ax}Bigr]_0^X-int_0^Xfrac{-1}{a}e^{-ax}dx$$



        $$frac{Xe^{-aX}}{a}+frac{1}{a}Bigl[frac{-1}{a}e^{-ax}Bigr]_0^X=$$



        $$frac{Xe^{-aX}}{a}+frac{1}{a^2}-frac{e^{-aX}}{a^2}$$



        now take the limit when $Xto+infty$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 18:44









        hamam_Abdallahhamam_Abdallah

        38.1k21634




        38.1k21634






























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