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A better lower bound of the definite integral $int_0^1fracln^2xe^2x,dx$
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In this post of my blog, I proved that $$int_0^1frac(x^2-3x+1)ln xe^x,dx=-frac1e.$$ Now if we apply the Cauchy-Schwarz inequality for integrals, we get $$frac1e^2leint_0^1(x^2-3x+1)^2,dxint_0^1fracln^2xe^2x,dx.$$ The first integral is easy to evaluate, so we arrive at $$int_0^1fracln^2xe^2x,dxgefrac3011e^2approx0.36909cdots$$ If I plug the integral into WA, the actual value is $2,_3F_3(1,1,1;2,2,2;-2)approx1.61511$ which involves hypergeometric functions. We can see that my approximation is not very useful.
So could a closer lower bound be found using 'simple' methods (e.g. not requiring the use of hypergeometric or gamma functions)?
P.S. An analytical method would be best.
integration definite-integrals cauchy-schwarz-inequality upper-lower-bounds
asked Jul 20 at 15:02
TheSimpliFire
9,67461951
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up vote
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In this post of my blog, I proved that $$int_0^1frac(x^2-3x+1)ln xe^x,dx=-frac1e.$$ Now if we apply the Cauchy-Schwarz inequality for integrals, we get $$frac1e^2leint_0^1(x^2-3x+1)^2,dxint_0^1fracln^2xe^2x,dx.$$ The first integral is easy to evaluate, so we arrive at $$int_0^1fracln^2xe^2x,dxgefrac3011e^2approx0.36909cdots$$ If I plug the integral into WA, the actual value is $2,_3F_3(1,1,1;2,2,2;-2)approx1.61511$ which involves hypergeometric functions. We can see that my approximation is not very useful.
So could a closer lower bound be found using 'simple' methods (e.g. not requiring the use of hypergeometric or gamma functions)?
P.S. An analytical method would be best.
integration definite-integrals cauchy-schwarz-inequality upper-lower-bounds
asked Jul 20 at 15:02
TheSimpliFire
9,67461951
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
In this post of my blog, I proved that $$int_0^1frac(x^2-3x+1)ln xe^x,dx=-frac1e.$$ Now if we apply the Cauchy-Schwarz inequality for integrals, we get $$frac1e^2leint_0^1(x^2-3x+1)^2,dxint_0^1fracln^2xe^2x,dx.$$ The first integral is easy to evaluate, so we arrive at $$int_0^1fracln^2xe^2x,dxgefrac3011e^2approx0.36909cdots$$ If I plug the integral into WA, the actual value is $2,_3F_3(1,1,1;2,2,2;-2)approx1.61511$ which involves hypergeometric functions. We can see that my approximation is not very useful.
So could a closer lower bound be found using 'simple' methods (e.g. not requiring the use of hypergeometric or gamma functions)?
P.S. An analytical method would be best.
integration definite-integrals cauchy-schwarz-inequality upper-lower-bounds
asked Jul 20 at 15:02
TheSimpliFire
9,67461951
In this post of my blog, I proved that $$int_0^1frac(x^2-3x+1)ln xe^x,dx=-frac1e.$$ Now if we apply the Cauchy-Schwarz inequality for integrals, we get $$frac1e^2leint_0^1(x^2-3x+1)^2,dxint_0^1fracln^2xe^2x,dx.$$ The first integral is easy to evaluate, so we arrive at $$int_0^1fracln^2xe^2x,dxgefrac3011e^2approx0.36909cdots$$ If I plug the integral into WA, the actual value is $2,_3F_3(1,1,1;2,2,2;-2)approx1.61511$ which involves hypergeometric functions. We can see that my approximation is not very useful.
So could a closer lower bound be found using 'simple' methods (e.g. not requiring the use of hypergeometric or gamma functions)?
P.S. An analytical method would be best.
integration definite-integrals cauchy-schwarz-inequality upper-lower-bounds
asked Jul 20 at 15:02
TheSimpliFire
9,67461951
edited Jul 20 at 15:31
asked Jul 20 at 15:02
TheSimpliFire
9,67461951
asked Jul 20 at 15:02
TheSimpliFire
9,67461951
asked Jul 20 at 15:02
TheSimpliFire
9,67461951
9,67461951
add a comment |Â
add a comment |Â
2 Answers
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You can just truncate the exponential series at an odd index: For $n in mathbbN$ we have
$$ I equiv int limits_0^1 ln^2(x) mathrme^-2 x , mathrmd x geq sum limits_k=0^2n-1 frac(-2)^kk! int limits_0^1 ln^2 (x) x^k , mathrmdx = sum limits_k=0^2n-1 frac(-1)^k 2^k+1k! (k+1)^3 equiv I_n , . $$
Then $I_1 = frac32 = 1.5$, $I_2 = frac347216 approx 1.60648$, $I_3 = frac13078981000 approx 1.61468$ and so on.
answered Jul 20 at 18:51
ComplexYetTrivial
2,607624
Thanks! I should have thought of Taylor series...
– TheSimpliFire
Jul 20 at 18:53
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up vote
0
down vote
It's trivial to get a better lower bound using a trivial method:
$$frac12 cdot 0.3 cdot 5 = 0.75$$
answered Jul 20 at 15:15
orlp
6,6051228
How could this be done analytically (without the aid of graphs)?
– TheSimpliFire
Jul 20 at 15:30
@TheSimpliFire You can show that the function is monotonic and then draw such a triangle below the tangent of the function.
– orlp
Jul 20 at 15:33
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You can just truncate the exponential series at an odd index: For $n in mathbbN$ we have
$$ I equiv int limits_0^1 ln^2(x) mathrme^-2 x , mathrmd x geq sum limits_k=0^2n-1 frac(-2)^kk! int limits_0^1 ln^2 (x) x^k , mathrmdx = sum limits_k=0^2n-1 frac(-1)^k 2^k+1k! (k+1)^3 equiv I_n , . $$
Then $I_1 = frac32 = 1.5$, $I_2 = frac347216 approx 1.60648$, $I_3 = frac13078981000 approx 1.61468$ and so on.
answered Jul 20 at 18:51
ComplexYetTrivial
2,607624
Thanks! I should have thought of Taylor series...
– TheSimpliFire
Jul 20 at 18:53
add a comment |Â
up vote
3
down vote
accepted
You can just truncate the exponential series at an odd index: For $n in mathbbN$ we have
$$ I equiv int limits_0^1 ln^2(x) mathrme^-2 x , mathrmd x geq sum limits_k=0^2n-1 frac(-2)^kk! int limits_0^1 ln^2 (x) x^k , mathrmdx = sum limits_k=0^2n-1 frac(-1)^k 2^k+1k! (k+1)^3 equiv I_n , . $$
Then $I_1 = frac32 = 1.5$, $I_2 = frac347216 approx 1.60648$, $I_3 = frac13078981000 approx 1.61468$ and so on.
answered Jul 20 at 18:51
ComplexYetTrivial
2,607624
Thanks! I should have thought of Taylor series...
– TheSimpliFire
Jul 20 at 18:53
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You can just truncate the exponential series at an odd index: For $n in mathbbN$ we have
$$ I equiv int limits_0^1 ln^2(x) mathrme^-2 x , mathrmd x geq sum limits_k=0^2n-1 frac(-2)^kk! int limits_0^1 ln^2 (x) x^k , mathrmdx = sum limits_k=0^2n-1 frac(-1)^k 2^k+1k! (k+1)^3 equiv I_n , . $$
Then $I_1 = frac32 = 1.5$, $I_2 = frac347216 approx 1.60648$, $I_3 = frac13078981000 approx 1.61468$ and so on.
answered Jul 20 at 18:51
ComplexYetTrivial
2,607624
You can just truncate the exponential series at an odd index: For $n in mathbbN$ we have
$$ I equiv int limits_0^1 ln^2(x) mathrme^-2 x , mathrmd x geq sum limits_k=0^2n-1 frac(-2)^kk! int limits_0^1 ln^2 (x) x^k , mathrmdx = sum limits_k=0^2n-1 frac(-1)^k 2^k+1k! (k+1)^3 equiv I_n , . $$
Then $I_1 = frac32 = 1.5$, $I_2 = frac347216 approx 1.60648$, $I_3 = frac13078981000 approx 1.61468$ and so on.
answered Jul 20 at 18:51
ComplexYetTrivial
2,607624
answered Jul 20 at 18:51
ComplexYetTrivial
2,607624
answered Jul 20 at 18:51
ComplexYetTrivial
2,607624
answered Jul 20 at 18:51
ComplexYetTrivial
2,607624
2,607624
Thanks! I should have thought of Taylor series...
– TheSimpliFire
Jul 20 at 18:53
add a comment |Â
Thanks! I should have thought of Taylor series...
– TheSimpliFire
Jul 20 at 18:53
Thanks! I should have thought of Taylor series...
– TheSimpliFire
Jul 20 at 18:53
Thanks! I should have thought of Taylor series...
– TheSimpliFire
Jul 20 at 18:53
add a comment |Â
up vote
0
down vote
It's trivial to get a better lower bound using a trivial method:
$$frac12 cdot 0.3 cdot 5 = 0.75$$
answered Jul 20 at 15:15
orlp
6,6051228
How could this be done analytically (without the aid of graphs)?
– TheSimpliFire
Jul 20 at 15:30
@TheSimpliFire You can show that the function is monotonic and then draw such a triangle below the tangent of the function.
– orlp
Jul 20 at 15:33
add a comment |Â
up vote
0
down vote
It's trivial to get a better lower bound using a trivial method:
$$frac12 cdot 0.3 cdot 5 = 0.75$$
answered Jul 20 at 15:15
orlp
6,6051228
How could this be done analytically (without the aid of graphs)?
– TheSimpliFire
Jul 20 at 15:30
@TheSimpliFire You can show that the function is monotonic and then draw such a triangle below the tangent of the function.
– orlp
Jul 20 at 15:33
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It's trivial to get a better lower bound using a trivial method:
$$frac12 cdot 0.3 cdot 5 = 0.75$$
answered Jul 20 at 15:15
orlp
6,6051228
It's trivial to get a better lower bound using a trivial method:
$$frac12 cdot 0.3 cdot 5 = 0.75$$
answered Jul 20 at 15:15
orlp
6,6051228
answered Jul 20 at 15:15
orlp
6,6051228
answered Jul 20 at 15:15
orlp
6,6051228
answered Jul 20 at 15:15
orlp
6,6051228
6,6051228
How could this be done analytically (without the aid of graphs)?
– TheSimpliFire
Jul 20 at 15:30
@TheSimpliFire You can show that the function is monotonic and then draw such a triangle below the tangent of the function.
– orlp
Jul 20 at 15:33
add a comment |Â
How could this be done analytically (without the aid of graphs)?
– TheSimpliFire
Jul 20 at 15:30
@TheSimpliFire You can show that the function is monotonic and then draw such a triangle below the tangent of the function.
– orlp
Jul 20 at 15:33
How could this be done analytically (without the aid of graphs)?
– TheSimpliFire
Jul 20 at 15:30
How could this be done analytically (without the aid of graphs)?
– TheSimpliFire
Jul 20 at 15:30
@TheSimpliFire You can show that the function is monotonic and then draw such a triangle below the tangent of the function.
– orlp
Jul 20 at 15:33
@TheSimpliFire You can show that the function is monotonic and then draw such a triangle below the tangent of the function.
– orlp
Jul 20 at 15:33
add a comment |Â
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