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Work out $int_0^pi/2cos^2n(x)lnsin(x) dx$
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$$F(n)=large int_0^pi/2cos^2n(x)lnsin(x)mathrm dx$$
$$cos^2n(x)=frac12^2n2n choose n+frac12^2n-1sum_k=0^n-12n choose kcosleft[2left(n-kright)xright]$$
$$I=frac12^2n2n choose nint_0^pi/2lnsin(x)mathrm dx+frac12^2n-1sum_k=0^n-12n choose kint_0^pi/2cosleft[2left(n-kright)xright]lnsin(x)mathrm dx$$
$$I=-fracpi ln(2)2^2n+12n choose n+frac12^2n-1sum_k=0^n-12n choose kint_0^pi/2cosleft[2left(n-kright)xright]lnsin(x)mathrm dx$$
I can't continue.
I would like to evaluate $F(n)$
calculus integration definite-integrals trigonometric-integrals
asked Jul 27 at 7:53
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51110
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up vote
3
down vote
favorite
$$F(n)=large int_0^pi/2cos^2n(x)lnsin(x)mathrm dx$$
$$cos^2n(x)=frac12^2n2n choose n+frac12^2n-1sum_k=0^n-12n choose kcosleft[2left(n-kright)xright]$$
$$I=frac12^2n2n choose nint_0^pi/2lnsin(x)mathrm dx+frac12^2n-1sum_k=0^n-12n choose kint_0^pi/2cosleft[2left(n-kright)xright]lnsin(x)mathrm dx$$
$$I=-fracpi ln(2)2^2n+12n choose n+frac12^2n-1sum_k=0^n-12n choose kint_0^pi/2cosleft[2left(n-kright)xright]lnsin(x)mathrm dx$$
I can't continue.
I would like to evaluate $F(n)$
calculus integration definite-integrals trigonometric-integrals
asked Jul 27 at 7:53
Bonjour
51110
Mathematica gives (for $n>-1/2$): $-fracsqrtpi Gamma left(n+frac12right) left(H_n+log (4)right)4 Gamma (n+1)$.
– David G. Stork
Jul 27 at 8:11
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$F(n)=large int_0^pi/2cos^2n(x)lnsin(x)mathrm dx$$
$$cos^2n(x)=frac12^2n2n choose n+frac12^2n-1sum_k=0^n-12n choose kcosleft[2left(n-kright)xright]$$
$$I=frac12^2n2n choose nint_0^pi/2lnsin(x)mathrm dx+frac12^2n-1sum_k=0^n-12n choose kint_0^pi/2cosleft[2left(n-kright)xright]lnsin(x)mathrm dx$$
$$I=-fracpi ln(2)2^2n+12n choose n+frac12^2n-1sum_k=0^n-12n choose kint_0^pi/2cosleft[2left(n-kright)xright]lnsin(x)mathrm dx$$
I can't continue.
I would like to evaluate $F(n)$
calculus integration definite-integrals trigonometric-integrals
asked Jul 27 at 7:53
Bonjour
51110
$$F(n)=large int_0^pi/2cos^2n(x)lnsin(x)mathrm dx$$
$$cos^2n(x)=frac12^2n2n choose n+frac12^2n-1sum_k=0^n-12n choose kcosleft[2left(n-kright)xright]$$
$$I=frac12^2n2n choose nint_0^pi/2lnsin(x)mathrm dx+frac12^2n-1sum_k=0^n-12n choose kint_0^pi/2cosleft[2left(n-kright)xright]lnsin(x)mathrm dx$$
$$I=-fracpi ln(2)2^2n+12n choose n+frac12^2n-1sum_k=0^n-12n choose kint_0^pi/2cosleft[2left(n-kright)xright]lnsin(x)mathrm dx$$
I can't continue.
I would like to evaluate $F(n)$
calculus integration definite-integrals trigonometric-integrals
asked Jul 27 at 7:53
Bonjour
51110
asked Jul 27 at 7:53
Bonjour
51110
asked Jul 27 at 7:53
Bonjour
51110
asked Jul 27 at 7:53
Bonjour
51110
51110
Mathematica gives (for $n>-1/2$): $-fracsqrtpi Gamma left(n+frac12right) left(H_n+log (4)right)4 Gamma (n+1)$.
– David G. Stork
Jul 27 at 8:11
add a comment |Â
Mathematica gives (for $n>-1/2$): $-fracsqrtpi Gamma left(n+frac12right) left(H_n+log (4)right)4 Gamma (n+1)$.
– David G. Stork
Jul 27 at 8:11
Mathematica gives (for $n>-1/2$): $-fracsqrtpi Gamma left(n+frac12right) left(H_n+log (4)right)4 Gamma (n+1)$.
– David G. Stork
Jul 27 at 8:11
Mathematica gives (for $n>-1/2$): $-fracsqrtpi Gamma left(n+frac12right) left(H_n+log (4)right)4 Gamma (n+1)$.
– David G. Stork
Jul 27 at 8:11
add a comment |Â
3 Answers
3
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oldest
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up vote
2
down vote
accepted
You can let $cos^2 (x) = t$ to obtain
beginalign
F(n) &= frac14 int limits_0^1 t^n-frac12 (1-t)^-frac12 ln(1-t) , mathrmd t \
&= frac14fracmathrmdmathrmds int limits_0^1 t^n-frac12 (1-t)^s - 1 , mathrmd t ~Bigg lvert_s=frac12 \
&= frac14 fracmathrmdmathrmds operatornameB left(n+frac12,sright)\
&= -fracGammaleft(n+frac12right) Gammaleft(frac12right)4Gamma(n+1) left[psi (n+1) - psi left(frac12right)right] , .
endalign
For $n in mathbbN_0$ the special values $Gammaleft(n+frac12right) = frac(2n)! sqrtpi4^n n!$ , $psi left(frac12right) = - gamma - 2 ln(2)$ and $psi(n+1) = H_n - gamma$ yield
$$F(n) = - fracpi4 frac(2n-1)!!(2n)!![H_n + 2 ln(2)] , . $$
answered Jul 27 at 8:33
ComplexYetTrivial
2,787624
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up vote
2
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Hint: With $beta$-function
$$beta(x,y) = 2int_0^pi/2sin^2x-1theta cos^2y-1theta dtheta = dfracGamma(x)Gamma(y)Gamma(x+y)$$
then
$$dfracddxbeta(x,n+frac12) =dfracddx 2int_0^pi/2sin^2x-1theta cos^2ntheta dtheta=int_0^pi/2sin^2x-1theta cos^2ntheta lnsin x dtheta$$
and
$$F(n)=left(dfracddxbeta(x,n+frac12)right)_x=frac12 =left(dfracddxdfracGamma(x)Gamma(n+frac12)Gamma(x+n+frac12)right)_x=frac12
$$
answered Jul 27 at 8:23
Nosrati
19.2k41544
This is elegant, for sure ! $to +1$
– Claude Leibovici
Jul 27 at 8:34
add a comment |Â
up vote
2
down vote
Note that for $xin (0,pi/2]$,
$$ln(sin(x))=lnleft(frace^ix-e^-ix2iright)
=-ln(2)-sum_j=1^inftyfraccos(2jx)j.$$
Hence, for $0leq k<n$ and $jgeq 1$,
$$int_0^pi/2cos(2left(n-kright)x)cos(2jx) dx=begincases
pi/4 &textif $n-k=j$\
0&textotherwise
endcases.$$
Therefore, following your approach, we obtain that
$$beginalign*
F(n)&=-fracpi2^2n+1left(ln(2)binom2nn+sum_k=0^n-1binom2nkfrac1n-kright)\
&=-fracpi2^2n+2binom2nnleft(2ln(2)+H_nright)endalign*.$$
answered Jul 27 at 8:23
Robert Z
83.8k954122
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You can let $cos^2 (x) = t$ to obtain
beginalign
F(n) &= frac14 int limits_0^1 t^n-frac12 (1-t)^-frac12 ln(1-t) , mathrmd t \
&= frac14fracmathrmdmathrmds int limits_0^1 t^n-frac12 (1-t)^s - 1 , mathrmd t ~Bigg lvert_s=frac12 \
&= frac14 fracmathrmdmathrmds operatornameB left(n+frac12,sright)\
&= -fracGammaleft(n+frac12right) Gammaleft(frac12right)4Gamma(n+1) left[psi (n+1) - psi left(frac12right)right] , .
endalign
For $n in mathbbN_0$ the special values $Gammaleft(n+frac12right) = frac(2n)! sqrtpi4^n n!$ , $psi left(frac12right) = - gamma - 2 ln(2)$ and $psi(n+1) = H_n - gamma$ yield
$$F(n) = - fracpi4 frac(2n-1)!!(2n)!![H_n + 2 ln(2)] , . $$
answered Jul 27 at 8:33
ComplexYetTrivial
2,787624
add a comment |Â
up vote
2
down vote
accepted
You can let $cos^2 (x) = t$ to obtain
beginalign
F(n) &= frac14 int limits_0^1 t^n-frac12 (1-t)^-frac12 ln(1-t) , mathrmd t \
&= frac14fracmathrmdmathrmds int limits_0^1 t^n-frac12 (1-t)^s - 1 , mathrmd t ~Bigg lvert_s=frac12 \
&= frac14 fracmathrmdmathrmds operatornameB left(n+frac12,sright)\
&= -fracGammaleft(n+frac12right) Gammaleft(frac12right)4Gamma(n+1) left[psi (n+1) - psi left(frac12right)right] , .
endalign
For $n in mathbbN_0$ the special values $Gammaleft(n+frac12right) = frac(2n)! sqrtpi4^n n!$ , $psi left(frac12right) = - gamma - 2 ln(2)$ and $psi(n+1) = H_n - gamma$ yield
$$F(n) = - fracpi4 frac(2n-1)!!(2n)!![H_n + 2 ln(2)] , . $$
answered Jul 27 at 8:33
ComplexYetTrivial
2,787624
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You can let $cos^2 (x) = t$ to obtain
beginalign
F(n) &= frac14 int limits_0^1 t^n-frac12 (1-t)^-frac12 ln(1-t) , mathrmd t \
&= frac14fracmathrmdmathrmds int limits_0^1 t^n-frac12 (1-t)^s - 1 , mathrmd t ~Bigg lvert_s=frac12 \
&= frac14 fracmathrmdmathrmds operatornameB left(n+frac12,sright)\
&= -fracGammaleft(n+frac12right) Gammaleft(frac12right)4Gamma(n+1) left[psi (n+1) - psi left(frac12right)right] , .
endalign
For $n in mathbbN_0$ the special values $Gammaleft(n+frac12right) = frac(2n)! sqrtpi4^n n!$ , $psi left(frac12right) = - gamma - 2 ln(2)$ and $psi(n+1) = H_n - gamma$ yield
$$F(n) = - fracpi4 frac(2n-1)!!(2n)!![H_n + 2 ln(2)] , . $$
answered Jul 27 at 8:33
ComplexYetTrivial
2,787624
You can let $cos^2 (x) = t$ to obtain
beginalign
F(n) &= frac14 int limits_0^1 t^n-frac12 (1-t)^-frac12 ln(1-t) , mathrmd t \
&= frac14fracmathrmdmathrmds int limits_0^1 t^n-frac12 (1-t)^s - 1 , mathrmd t ~Bigg lvert_s=frac12 \
&= frac14 fracmathrmdmathrmds operatornameB left(n+frac12,sright)\
&= -fracGammaleft(n+frac12right) Gammaleft(frac12right)4Gamma(n+1) left[psi (n+1) - psi left(frac12right)right] , .
endalign
For $n in mathbbN_0$ the special values $Gammaleft(n+frac12right) = frac(2n)! sqrtpi4^n n!$ , $psi left(frac12right) = - gamma - 2 ln(2)$ and $psi(n+1) = H_n - gamma$ yield
$$F(n) = - fracpi4 frac(2n-1)!!(2n)!![H_n + 2 ln(2)] , . $$
answered Jul 27 at 8:33
ComplexYetTrivial
2,787624
answered Jul 27 at 8:33
ComplexYetTrivial
2,787624
answered Jul 27 at 8:33
ComplexYetTrivial
2,787624
answered Jul 27 at 8:33
ComplexYetTrivial
2,787624
2,787624
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint: With $beta$-function
$$beta(x,y) = 2int_0^pi/2sin^2x-1theta cos^2y-1theta dtheta = dfracGamma(x)Gamma(y)Gamma(x+y)$$
then
$$dfracddxbeta(x,n+frac12) =dfracddx 2int_0^pi/2sin^2x-1theta cos^2ntheta dtheta=int_0^pi/2sin^2x-1theta cos^2ntheta lnsin x dtheta$$
and
$$F(n)=left(dfracddxbeta(x,n+frac12)right)_x=frac12 =left(dfracddxdfracGamma(x)Gamma(n+frac12)Gamma(x+n+frac12)right)_x=frac12
$$
answered Jul 27 at 8:23
Nosrati
19.2k41544
This is elegant, for sure ! $to +1$
– Claude Leibovici
Jul 27 at 8:34
add a comment |Â
up vote
2
down vote
Hint: With $beta$-function
$$beta(x,y) = 2int_0^pi/2sin^2x-1theta cos^2y-1theta dtheta = dfracGamma(x)Gamma(y)Gamma(x+y)$$
then
$$dfracddxbeta(x,n+frac12) =dfracddx 2int_0^pi/2sin^2x-1theta cos^2ntheta dtheta=int_0^pi/2sin^2x-1theta cos^2ntheta lnsin x dtheta$$
and
$$F(n)=left(dfracddxbeta(x,n+frac12)right)_x=frac12 =left(dfracddxdfracGamma(x)Gamma(n+frac12)Gamma(x+n+frac12)right)_x=frac12
$$
answered Jul 27 at 8:23
Nosrati
19.2k41544
This is elegant, for sure ! $to +1$
– Claude Leibovici
Jul 27 at 8:34
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: With $beta$-function
$$beta(x,y) = 2int_0^pi/2sin^2x-1theta cos^2y-1theta dtheta = dfracGamma(x)Gamma(y)Gamma(x+y)$$
then
$$dfracddxbeta(x,n+frac12) =dfracddx 2int_0^pi/2sin^2x-1theta cos^2ntheta dtheta=int_0^pi/2sin^2x-1theta cos^2ntheta lnsin x dtheta$$
and
$$F(n)=left(dfracddxbeta(x,n+frac12)right)_x=frac12 =left(dfracddxdfracGamma(x)Gamma(n+frac12)Gamma(x+n+frac12)right)_x=frac12
$$
answered Jul 27 at 8:23
Nosrati
19.2k41544
Hint: With $beta$-function
$$beta(x,y) = 2int_0^pi/2sin^2x-1theta cos^2y-1theta dtheta = dfracGamma(x)Gamma(y)Gamma(x+y)$$
then
$$dfracddxbeta(x,n+frac12) =dfracddx 2int_0^pi/2sin^2x-1theta cos^2ntheta dtheta=int_0^pi/2sin^2x-1theta cos^2ntheta lnsin x dtheta$$
and
$$F(n)=left(dfracddxbeta(x,n+frac12)right)_x=frac12 =left(dfracddxdfracGamma(x)Gamma(n+frac12)Gamma(x+n+frac12)right)_x=frac12
$$
answered Jul 27 at 8:23
Nosrati
19.2k41544
answered Jul 27 at 8:23
Nosrati
19.2k41544
answered Jul 27 at 8:23
Nosrati
19.2k41544
answered Jul 27 at 8:23
Nosrati
19.2k41544
19.2k41544
This is elegant, for sure ! $to +1$
– Claude Leibovici
Jul 27 at 8:34
add a comment |Â
This is elegant, for sure ! $to +1$
– Claude Leibovici
Jul 27 at 8:34
This is elegant, for sure ! $to +1$
– Claude Leibovici
Jul 27 at 8:34
This is elegant, for sure ! $to +1$
– Claude Leibovici
Jul 27 at 8:34
add a comment |Â
up vote
2
down vote
Note that for $xin (0,pi/2]$,
$$ln(sin(x))=lnleft(frace^ix-e^-ix2iright)
=-ln(2)-sum_j=1^inftyfraccos(2jx)j.$$
Hence, for $0leq k<n$ and $jgeq 1$,
$$int_0^pi/2cos(2left(n-kright)x)cos(2jx) dx=begincases
pi/4 &textif $n-k=j$\
0&textotherwise
endcases.$$
Therefore, following your approach, we obtain that
$$beginalign*
F(n)&=-fracpi2^2n+1left(ln(2)binom2nn+sum_k=0^n-1binom2nkfrac1n-kright)\
&=-fracpi2^2n+2binom2nnleft(2ln(2)+H_nright)endalign*.$$
answered Jul 27 at 8:23
Robert Z
83.8k954122
add a comment |Â
up vote
2
down vote
Note that for $xin (0,pi/2]$,
$$ln(sin(x))=lnleft(frace^ix-e^-ix2iright)
=-ln(2)-sum_j=1^inftyfraccos(2jx)j.$$
Hence, for $0leq k<n$ and $jgeq 1$,
$$int_0^pi/2cos(2left(n-kright)x)cos(2jx) dx=begincases
pi/4 &textif $n-k=j$\
0&textotherwise
endcases.$$
Therefore, following your approach, we obtain that
$$beginalign*
F(n)&=-fracpi2^2n+1left(ln(2)binom2nn+sum_k=0^n-1binom2nkfrac1n-kright)\
&=-fracpi2^2n+2binom2nnleft(2ln(2)+H_nright)endalign*.$$
answered Jul 27 at 8:23
Robert Z
83.8k954122
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Note that for $xin (0,pi/2]$,
$$ln(sin(x))=lnleft(frace^ix-e^-ix2iright)
=-ln(2)-sum_j=1^inftyfraccos(2jx)j.$$
Hence, for $0leq k<n$ and $jgeq 1$,
$$int_0^pi/2cos(2left(n-kright)x)cos(2jx) dx=begincases
pi/4 &textif $n-k=j$\
0&textotherwise
endcases.$$
Therefore, following your approach, we obtain that
$$beginalign*
F(n)&=-fracpi2^2n+1left(ln(2)binom2nn+sum_k=0^n-1binom2nkfrac1n-kright)\
&=-fracpi2^2n+2binom2nnleft(2ln(2)+H_nright)endalign*.$$
answered Jul 27 at 8:23
Robert Z
83.8k954122
Note that for $xin (0,pi/2]$,
$$ln(sin(x))=lnleft(frace^ix-e^-ix2iright)
=-ln(2)-sum_j=1^inftyfraccos(2jx)j.$$
Hence, for $0leq k<n$ and $jgeq 1$,
$$int_0^pi/2cos(2left(n-kright)x)cos(2jx) dx=begincases
pi/4 &textif $n-k=j$\
0&textotherwise
endcases.$$
Therefore, following your approach, we obtain that
$$beginalign*
F(n)&=-fracpi2^2n+1left(ln(2)binom2nn+sum_k=0^n-1binom2nkfrac1n-kright)\
&=-fracpi2^2n+2binom2nnleft(2ln(2)+H_nright)endalign*.$$
answered Jul 27 at 8:23
Robert Z
83.8k954122
edited Jul 27 at 8:51
answered Jul 27 at 8:23
Robert Z
83.8k954122
answered Jul 27 at 8:23
Robert Z
83.8k954122
answered Jul 27 at 8:23
Robert Z
83.8k954122
83.8k954122
add a comment |Â
add a comment |Â
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Mathematica gives (for $n>-1/2$): $-fracsqrtpi Gamma left(n+frac12right) left(H_n+log (4)right)4 Gamma (n+1)$.
– David G. Stork
Jul 27 at 8:11