definite integration - solution breakdown
$begingroup$
$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$
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
asked Dec 12 '18 at 18:25
user1607user1607
1718
$endgroup$
add a comment |
$begingroup$
$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$
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
asked Dec 12 '18 at 18:25
user1607user1607
1718
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$
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
asked Dec 12 '18 at 18:25
user1607user1607
1718
$endgroup$
$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$
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
asked Dec 12 '18 at 18:25
user1607user1607
1718
asked Dec 12 '18 at 18:25
user1607user1607
1718
edited Dec 12 '18 at 18:48
user1607
asked Dec 12 '18 at 18:25
user1607user1607
1718
asked Dec 12 '18 at 18:25
user1607user1607
1718
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
add a comment |
$begingroup$
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
add a comment |
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 $Xto+infty$.
answered Dec 12 '18 at 18:44
hamam_Abdallahhamam_Abdallah
38.1k21634
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add a comment |
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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 $Xto+infty$.
answered Dec 12 '18 at 18:44
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
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 $Xto+infty$.
answered Dec 12 '18 at 18:44
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
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 $Xto+infty$.
answered Dec 12 '18 at 18:44
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
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 $Xto+infty$.
answered Dec 12 '18 at 18:44
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 12 '18 at 18:44
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 12 '18 at 18:44
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Dec 12 '18 at 18:44
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
add a comment |
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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