Mixing
integral by part with distribution function.
Clash Royale CLAN TAG#URR8PPP
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I have a solution for my homework and I was not sure how the integral by part was implemented in the following equations... Could anyone give explanation to hidden steps that lead to the equations below? thank you!
integration
edited Jul 15 at 2:50
Parcly Taxel
33.6k136588
asked Jul 15 at 2:21
David Kim
83
add a comment |Â
up vote
0
down vote
favorite
I have a solution for my homework and I was not sure how the integral by part was implemented in the following equations... Could anyone give explanation to hidden steps that lead to the equations below? thank you!
integration
edited Jul 15 at 2:50
Parcly Taxel
33.6k136588
asked Jul 15 at 2:21
David Kim
83
Please tell us what you did and why you are confused. You will see that people will put effort into their response if you put effort into your question.
– bkarthik
Jul 15 at 2:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a solution for my homework and I was not sure how the integral by part was implemented in the following equations... Could anyone give explanation to hidden steps that lead to the equations below? thank you!
integration
edited Jul 15 at 2:50
Parcly Taxel
33.6k136588
asked Jul 15 at 2:21
David Kim
83
I have a solution for my homework and I was not sure how the integral by part was implemented in the following equations... Could anyone give explanation to hidden steps that lead to the equations below? thank you!
integration
edited Jul 15 at 2:50
Parcly Taxel
33.6k136588
asked Jul 15 at 2:21
David Kim
83
edited Jul 15 at 2:50
Parcly Taxel
33.6k136588
edited Jul 15 at 2:50
Parcly Taxel
33.6k136588
edited Jul 15 at 2:50
Parcly Taxel
33.6k136588
33.6k136588
asked Jul 15 at 2:21
David Kim
83
asked Jul 15 at 2:21
David Kim
83
asked Jul 15 at 2:21
David Kim
83
83
Please tell us what you did and why you are confused. You will see that people will put effort into their response if you put effort into your question.
– bkarthik
Jul 15 at 2:37
add a comment |Â
Please tell us what you did and why you are confused. You will see that people will put effort into their response if you put effort into your question.
– bkarthik
Jul 15 at 2:37
Please tell us what you did and why you are confused. You will see that people will put effort into their response if you put effort into your question.
– bkarthik
Jul 15 at 2:37
Please tell us what you did and why you are confused. You will see that people will put effort into their response if you put effort into your question.
– bkarthik
Jul 15 at 2:37
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
From differential you know that for every two differentiable functions $u$ and $v$ :
$$d(uv)=du.v+dv.u$$
or
$$udv=d(uv)-vdu$$
integration gives
beginalign
int udv
&=int d(uv)-int vdu\
&=uv-int vdu
endalign
this formula uses in integration by parts. Here we have $displaystyle H(p)=int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))$ and then let $u=theta$, $dv=ddfracF(theta)F(r^-1(p))$ or $v=dfracF(theta)F(r^-1(p))$ hence
beginalign
int udv
&=int d(uv)-int vdu\
int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))
&=theta~dfracF(theta)F(r^-1(p))-int dfracF(theta)F(r^-1(p)) dtheta
endalign
set upper and lower bounds then
beginalign
int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))
&=theta~dfracF(theta)F(r^-1(p))Big|_theta^r^-1(p)-int_theta^r^-1(p) dfracF(theta)F(r^-1(p)) dtheta
endalign
answered Jul 15 at 4:04
Nosrati
19.9k41644
Thank you very much for your help!
– David Kim
Jul 15 at 14:12
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
From differential you know that for every two differentiable functions $u$ and $v$ :
$$d(uv)=du.v+dv.u$$
or
$$udv=d(uv)-vdu$$
integration gives
beginalign
int udv
&=int d(uv)-int vdu\
&=uv-int vdu
endalign
this formula uses in integration by parts. Here we have $displaystyle H(p)=int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))$ and then let $u=theta$, $dv=ddfracF(theta)F(r^-1(p))$ or $v=dfracF(theta)F(r^-1(p))$ hence
beginalign
int udv
&=int d(uv)-int vdu\
int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))
&=theta~dfracF(theta)F(r^-1(p))-int dfracF(theta)F(r^-1(p)) dtheta
endalign
set upper and lower bounds then
beginalign
int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))
&=theta~dfracF(theta)F(r^-1(p))Big|_theta^r^-1(p)-int_theta^r^-1(p) dfracF(theta)F(r^-1(p)) dtheta
endalign
answered Jul 15 at 4:04
Nosrati
19.9k41644
Thank you very much for your help!
– David Kim
Jul 15 at 14:12
add a comment |Â
up vote
0
down vote
accepted
From differential you know that for every two differentiable functions $u$ and $v$ :
$$d(uv)=du.v+dv.u$$
or
$$udv=d(uv)-vdu$$
integration gives
beginalign
int udv
&=int d(uv)-int vdu\
&=uv-int vdu
endalign
this formula uses in integration by parts. Here we have $displaystyle H(p)=int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))$ and then let $u=theta$, $dv=ddfracF(theta)F(r^-1(p))$ or $v=dfracF(theta)F(r^-1(p))$ hence
beginalign
int udv
&=int d(uv)-int vdu\
int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))
&=theta~dfracF(theta)F(r^-1(p))-int dfracF(theta)F(r^-1(p)) dtheta
endalign
set upper and lower bounds then
beginalign
int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))
&=theta~dfracF(theta)F(r^-1(p))Big|_theta^r^-1(p)-int_theta^r^-1(p) dfracF(theta)F(r^-1(p)) dtheta
endalign
answered Jul 15 at 4:04
Nosrati
19.9k41644
Thank you very much for your help!
– David Kim
Jul 15 at 14:12
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
From differential you know that for every two differentiable functions $u$ and $v$ :
$$d(uv)=du.v+dv.u$$
or
$$udv=d(uv)-vdu$$
integration gives
beginalign
int udv
&=int d(uv)-int vdu\
&=uv-int vdu
endalign
this formula uses in integration by parts. Here we have $displaystyle H(p)=int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))$ and then let $u=theta$, $dv=ddfracF(theta)F(r^-1(p))$ or $v=dfracF(theta)F(r^-1(p))$ hence
beginalign
int udv
&=int d(uv)-int vdu\
int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))
&=theta~dfracF(theta)F(r^-1(p))-int dfracF(theta)F(r^-1(p)) dtheta
endalign
set upper and lower bounds then
beginalign
int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))
&=theta~dfracF(theta)F(r^-1(p))Big|_theta^r^-1(p)-int_theta^r^-1(p) dfracF(theta)F(r^-1(p)) dtheta
endalign
answered Jul 15 at 4:04
Nosrati
19.9k41644
From differential you know that for every two differentiable functions $u$ and $v$ :
$$d(uv)=du.v+dv.u$$
or
$$udv=d(uv)-vdu$$
integration gives
beginalign
int udv
&=int d(uv)-int vdu\
&=uv-int vdu
endalign
this formula uses in integration by parts. Here we have $displaystyle H(p)=int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))$ and then let $u=theta$, $dv=ddfracF(theta)F(r^-1(p))$ or $v=dfracF(theta)F(r^-1(p))$ hence
beginalign
int udv
&=int d(uv)-int vdu\
int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))
&=theta~dfracF(theta)F(r^-1(p))-int dfracF(theta)F(r^-1(p)) dtheta
endalign
set upper and lower bounds then
beginalign
int_theta^r^-1(p)theta ddfracF(theta)F(r^-1(p))
&=theta~dfracF(theta)F(r^-1(p))Big|_theta^r^-1(p)-int_theta^r^-1(p) dfracF(theta)F(r^-1(p)) dtheta
endalign
answered Jul 15 at 4:04
Nosrati
19.9k41644
answered Jul 15 at 4:04
Nosrati
19.9k41644
answered Jul 15 at 4:04
Nosrati
19.9k41644
answered Jul 15 at 4:04
Nosrati
19.9k41644
19.9k41644
Thank you very much for your help!
– David Kim
Jul 15 at 14:12
add a comment |Â
Thank you very much for your help!
– David Kim
Jul 15 at 14:12
Thank you very much for your help!
– David Kim
Jul 15 at 14:12
Thank you very much for your help!
– David Kim
Jul 15 at 14:12
add a comment |Â
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Please tell us what you did and why you are confused. You will see that people will put effort into their response if you put effort into your question.
– bkarthik
Jul 15 at 2:37