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Evaluate this integral: $int_y_min^infty (1 - ty)^1-alphay^-betady$, $alpha, beta in (2, 3)$
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This problem is a different approach to my earlier question here
The core integral $int_y_min^infty (1 - ty)^1-alphay^-betady$ resembles the Beta function, but I am unable to get to the substitution to bring it to the correct form.
Applying integration by parts with the notation $int udv = uv - int vdu$, where $u = (1-ty)^1-alpha$ and $v = dfracy^-beta1-beta$ gets me back to an integral that is similar to the original.
Applying the transformation: $ty = z$, we get $int_ty_min^infty (1 - z)^1-alphaz^-betadz$, but I am unable to reduce this further with $alpha, beta in (2, 3)$
Any hints?
calculus integration
asked Jul 21 at 3:53
buzaku
3091212
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up vote
0
down vote
favorite
This problem is a different approach to my earlier question here
The core integral $int_y_min^infty (1 - ty)^1-alphay^-betady$ resembles the Beta function, but I am unable to get to the substitution to bring it to the correct form.
Applying integration by parts with the notation $int udv = uv - int vdu$, where $u = (1-ty)^1-alpha$ and $v = dfracy^-beta1-beta$ gets me back to an integral that is similar to the original.
Applying the transformation: $ty = z$, we get $int_ty_min^infty (1 - z)^1-alphaz^-betadz$, but I am unable to reduce this further with $alpha, beta in (2, 3)$
Any hints?
calculus integration
asked Jul 21 at 3:53
buzaku
3091212
Try $z=1/y$. $$
– Szeto
Jul 21 at 4:28
Mathematica gives: $-e^-i pi alpha t^beta -1 B_frac1t textym(alpha +beta -2,2-alpha )$.
– David G. Stork
Jul 21 at 4:51
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This problem is a different approach to my earlier question here
The core integral $int_y_min^infty (1 - ty)^1-alphay^-betady$ resembles the Beta function, but I am unable to get to the substitution to bring it to the correct form.
Applying integration by parts with the notation $int udv = uv - int vdu$, where $u = (1-ty)^1-alpha$ and $v = dfracy^-beta1-beta$ gets me back to an integral that is similar to the original.
Applying the transformation: $ty = z$, we get $int_ty_min^infty (1 - z)^1-alphaz^-betadz$, but I am unable to reduce this further with $alpha, beta in (2, 3)$
Any hints?
calculus integration
asked Jul 21 at 3:53
buzaku
3091212
This problem is a different approach to my earlier question here
The core integral $int_y_min^infty (1 - ty)^1-alphay^-betady$ resembles the Beta function, but I am unable to get to the substitution to bring it to the correct form.
Applying integration by parts with the notation $int udv = uv - int vdu$, where $u = (1-ty)^1-alpha$ and $v = dfracy^-beta1-beta$ gets me back to an integral that is similar to the original.
Applying the transformation: $ty = z$, we get $int_ty_min^infty (1 - z)^1-alphaz^-betadz$, but I am unable to reduce this further with $alpha, beta in (2, 3)$
Any hints?
calculus integration
asked Jul 21 at 3:53
buzaku
3091212
asked Jul 21 at 3:53
buzaku
3091212
asked Jul 21 at 3:53
buzaku
3091212
asked Jul 21 at 3:53
buzaku
3091212
3091212
Try $z=1/y$. $$
– Szeto
Jul 21 at 4:28
Mathematica gives: $-e^-i pi alpha t^beta -1 B_frac1t textym(alpha +beta -2,2-alpha )$.
– David G. Stork
Jul 21 at 4:51
add a comment |Â
Try $z=1/y$. $$
– Szeto
Jul 21 at 4:28
Mathematica gives: $-e^-i pi alpha t^beta -1 B_frac1t textym(alpha +beta -2,2-alpha )$.
– David G. Stork
Jul 21 at 4:51
Try $z=1/y$. $$
– Szeto
Jul 21 at 4:28
Try $z=1/y$. $$
– Szeto
Jul 21 at 4:28
Mathematica gives: $-e^-i pi alpha t^beta -1 B_frac1t textym(alpha +beta -2,2-alpha )$.
– David G. Stork
Jul 21 at 4:51
Mathematica gives: $-e^-i pi alpha t^beta -1 B_frac1t textym(alpha +beta -2,2-alpha )$.
– David G. Stork
Jul 21 at 4:51
add a comment |Â
1 Answer
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By the substitution $z=frac1ty$, we get (let $m=y_textmin$)
$$beginalign
int^1/mt_0(1-frac1z)^1-alphat^beta-1z^beta-2dz & =t^beta-1int^1/mt_0(frac1-zz)^1-alphaz^beta-2dz\
& =t^beta-1int^1/mt_0(1-z)^1-alphaz^alpha+beta-3dz\
& =t^beta-1int^1/mt_0(e^pi i(1-z))^1-alphaz^alpha+beta-3dz \
& =t^beta-1e^pi i(1-alpha)int^1/mt_0(1-z)^1-alphaz^alpha+beta-3dz\
& =-t^beta-1e^-pi ialphaB(frac1mt;alpha+beta-2,2-alpha)
endalign$$
which is the same as Mathematica's result.
answered Jul 21 at 13:32
Szeto
4,1631521
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
By the substitution $z=frac1ty$, we get (let $m=y_textmin$)
$$beginalign
int^1/mt_0(1-frac1z)^1-alphat^beta-1z^beta-2dz & =t^beta-1int^1/mt_0(frac1-zz)^1-alphaz^beta-2dz\
& =t^beta-1int^1/mt_0(1-z)^1-alphaz^alpha+beta-3dz\
& =t^beta-1int^1/mt_0(e^pi i(1-z))^1-alphaz^alpha+beta-3dz \
& =t^beta-1e^pi i(1-alpha)int^1/mt_0(1-z)^1-alphaz^alpha+beta-3dz\
& =-t^beta-1e^-pi ialphaB(frac1mt;alpha+beta-2,2-alpha)
endalign$$
which is the same as Mathematica's result.
answered Jul 21 at 13:32
Szeto
4,1631521
add a comment |Â
up vote
0
down vote
accepted
By the substitution $z=frac1ty$, we get (let $m=y_textmin$)
$$beginalign
int^1/mt_0(1-frac1z)^1-alphat^beta-1z^beta-2dz & =t^beta-1int^1/mt_0(frac1-zz)^1-alphaz^beta-2dz\
& =t^beta-1int^1/mt_0(1-z)^1-alphaz^alpha+beta-3dz\
& =t^beta-1int^1/mt_0(e^pi i(1-z))^1-alphaz^alpha+beta-3dz \
& =t^beta-1e^pi i(1-alpha)int^1/mt_0(1-z)^1-alphaz^alpha+beta-3dz\
& =-t^beta-1e^-pi ialphaB(frac1mt;alpha+beta-2,2-alpha)
endalign$$
which is the same as Mathematica's result.
answered Jul 21 at 13:32
Szeto
4,1631521
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
By the substitution $z=frac1ty$, we get (let $m=y_textmin$)
$$beginalign
int^1/mt_0(1-frac1z)^1-alphat^beta-1z^beta-2dz & =t^beta-1int^1/mt_0(frac1-zz)^1-alphaz^beta-2dz\
& =t^beta-1int^1/mt_0(1-z)^1-alphaz^alpha+beta-3dz\
& =t^beta-1int^1/mt_0(e^pi i(1-z))^1-alphaz^alpha+beta-3dz \
& =t^beta-1e^pi i(1-alpha)int^1/mt_0(1-z)^1-alphaz^alpha+beta-3dz\
& =-t^beta-1e^-pi ialphaB(frac1mt;alpha+beta-2,2-alpha)
endalign$$
which is the same as Mathematica's result.
answered Jul 21 at 13:32
Szeto
4,1631521
By the substitution $z=frac1ty$, we get (let $m=y_textmin$)
$$beginalign
int^1/mt_0(1-frac1z)^1-alphat^beta-1z^beta-2dz & =t^beta-1int^1/mt_0(frac1-zz)^1-alphaz^beta-2dz\
& =t^beta-1int^1/mt_0(1-z)^1-alphaz^alpha+beta-3dz\
& =t^beta-1int^1/mt_0(e^pi i(1-z))^1-alphaz^alpha+beta-3dz \
& =t^beta-1e^pi i(1-alpha)int^1/mt_0(1-z)^1-alphaz^alpha+beta-3dz\
& =-t^beta-1e^-pi ialphaB(frac1mt;alpha+beta-2,2-alpha)
endalign$$
which is the same as Mathematica's result.
answered Jul 21 at 13:32
Szeto
4,1631521
answered Jul 21 at 13:32
Szeto
4,1631521
answered Jul 21 at 13:32
Szeto
4,1631521
answered Jul 21 at 13:32
Szeto
4,1631521
4,1631521
add a comment |Â
add a comment |Â
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Try $z=1/y$. $$
– Szeto
Jul 21 at 4:28
Mathematica gives: $-e^-i pi alpha t^beta -1 B_frac1t textym(alpha +beta -2,2-alpha )$.
– David G. Stork
Jul 21 at 4:51