An Integral from Irresistible Integrals
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I have been trying to work out a question from Irresistible Integrals, but I don't know how to proceed. The exact problem is Exercise 13.7.1. It asks us to show that $$zeta(3)=4-zeta(2)+frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt$$
The $zeta$ in question is the Riemann Zeta Function. The book does this integral through something that its calls its "Master Formula," but asks the reader to check it directly, which I assume means to compute the integral directly. The book gives the hint of expanding the integrand as a series, using $$frac{1}{sqrt{1-4x}}=sum_{n=0}^inftybinom{2n}{n}x^n$$
After applying this, I have $$int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=sum_{n=0}^inftybinom{2n}{n}int_0^{1/2}t^{2n+1}ln^3(t)dt$$
$$=-sum_{n=0}^inftyfrac{1}{2^{2n+2}}binom{2n}{n}left(frac{ln^3(2)}{2n+2}+frac{3ln^2(2)}{(2n+2)^2}+frac{6ln(2)}{(2n+2)^3}+frac{6}{(2n+2)^4}right)$$
(If someone wants to check that to makes sure I'm right, then by all means, go for it) I get stuck from here. I have a few tricks to evaluate the sum for the first two terms, but not the last two terms. If you have any ideas on how to evaluate this sum (or an alternate way of evaluating this integral in the first place), let me know.
Thanks!
Edit: Considering that this integral is being multiplied by $8/3$ in the original problem, it may be convenient to fact or out the $8/3$ to get
$$frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=$$
$$-sum_{n=0}^inftyfrac{1}{3cdot2^{2n-1}}binom{2n}{n}left(frac{ln^3(2)}{2n+2}+frac{3ln^2(2)}{(2n+2)^2}+frac{6ln(2)}{(2n+2)^3}+frac{6}{(2n+2)^4}right)$$
It also might be more convenient to factor the $2$'s from the denominators and get
$$frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=$$
$$-sum_{n=0}^inftyfrac{1}{3cdot2^{2n}}binom{2n}{n}left(frac{ln^3(2)}{n+1}+frac{3}{2}frac{ln^2(2)}{(n+1)^2}+frac{3}{2}frac{ln(2)}{(n+1)^3}+frac{3}{4}frac{1}{(n+1)^4}right)$$
It looks a bit nicer, but I still don't know how to proceed.
Edit 2: A good suggestion is to turn this integral into a form of the Beta function. This can be done a few different ways, but if we take $u=4t^2$ we get
$$int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=frac{1}{8}int_0^1(1-u)^{-1/2}ln^3left(sqrt{frac{u}{4}}right)du$$
$$=frac{1}{64}int_0^1(1-u)^{-1/2}left(ln^3(u)-3ln^2(u)ln(4)+3ln(u)ln^24)-ln^3(4)right)du$$
$$=frac{1}{64}left(frac{partial^3B}{partial x^3}(1,1/2)-3ln(4)frac{partial^2B}{partial x^2}(1,1/2)+3ln^2(4)frac{partial B}{partial x}(1,1/2)-ln^3(4)B(1,1/2)right)$$
At this point, derivatives of the Beta function are tricky, but they boil down to derivatives of the $psi$ function, which I am not very strong in. If someone has a solution from here, that would be handy as well, but I am hoping that someone can give a solution somehow using a series approach, as suggested in the text.
calculus integration sequences-and-series
asked Nov 4 at 4:05
Josh B.
7098
|
show 6 more comments
up vote
14
down vote
favorite
I have been trying to work out a question from Irresistible Integrals, but I don't know how to proceed. The exact problem is Exercise 13.7.1. It asks us to show that $$zeta(3)=4-zeta(2)+frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt$$
The $zeta$ in question is the Riemann Zeta Function. The book does this integral through something that its calls its "Master Formula," but asks the reader to check it directly, which I assume means to compute the integral directly. The book gives the hint of expanding the integrand as a series, using $$frac{1}{sqrt{1-4x}}=sum_{n=0}^inftybinom{2n}{n}x^n$$
After applying this, I have $$int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=sum_{n=0}^inftybinom{2n}{n}int_0^{1/2}t^{2n+1}ln^3(t)dt$$
$$=-sum_{n=0}^inftyfrac{1}{2^{2n+2}}binom{2n}{n}left(frac{ln^3(2)}{2n+2}+frac{3ln^2(2)}{(2n+2)^2}+frac{6ln(2)}{(2n+2)^3}+frac{6}{(2n+2)^4}right)$$
(If someone wants to check that to makes sure I'm right, then by all means, go for it) I get stuck from here. I have a few tricks to evaluate the sum for the first two terms, but not the last two terms. If you have any ideas on how to evaluate this sum (or an alternate way of evaluating this integral in the first place), let me know.
Thanks!
Edit: Considering that this integral is being multiplied by $8/3$ in the original problem, it may be convenient to fact or out the $8/3$ to get
$$frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=$$
$$-sum_{n=0}^inftyfrac{1}{3cdot2^{2n-1}}binom{2n}{n}left(frac{ln^3(2)}{2n+2}+frac{3ln^2(2)}{(2n+2)^2}+frac{6ln(2)}{(2n+2)^3}+frac{6}{(2n+2)^4}right)$$
It also might be more convenient to factor the $2$'s from the denominators and get
$$frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=$$
$$-sum_{n=0}^inftyfrac{1}{3cdot2^{2n}}binom{2n}{n}left(frac{ln^3(2)}{n+1}+frac{3}{2}frac{ln^2(2)}{(n+1)^2}+frac{3}{2}frac{ln(2)}{(n+1)^3}+frac{3}{4}frac{1}{(n+1)^4}right)$$
It looks a bit nicer, but I still don't know how to proceed.
Edit 2: A good suggestion is to turn this integral into a form of the Beta function. This can be done a few different ways, but if we take $u=4t^2$ we get
$$int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=frac{1}{8}int_0^1(1-u)^{-1/2}ln^3left(sqrt{frac{u}{4}}right)du$$
$$=frac{1}{64}int_0^1(1-u)^{-1/2}left(ln^3(u)-3ln^2(u)ln(4)+3ln(u)ln^24)-ln^3(4)right)du$$
$$=frac{1}{64}left(frac{partial^3B}{partial x^3}(1,1/2)-3ln(4)frac{partial^2B}{partial x^2}(1,1/2)+3ln^2(4)frac{partial B}{partial x}(1,1/2)-ln^3(4)B(1,1/2)right)$$
At this point, derivatives of the Beta function are tricky, but they boil down to derivatives of the $psi$ function, which I am not very strong in. If someone has a solution from here, that would be handy as well, but I am hoping that someone can give a solution somehow using a series approach, as suggested in the text.
calculus integration sequences-and-series
asked Nov 4 at 4:05
Josh B.
7098
Have you tried evaluating a few iterations of n and seeing if there is a pattern in cancellation?
– Henry Lee
Nov 4 at 11:27
I have tried writing out a few terms to look for a pattern, but there wasn't anything obvious to me. If I'm interpreting you right, then I would say there is no cancellation, since everything inside the final sum is positive.
– Josh B.
Nov 4 at 16:09
Yes that sounds correct, it seems that the only way to do it would be to split it into several different summations and try to find a general formula for each one. But I am not sure that one will exist
– Henry Lee
Nov 4 at 16:11
1
My main ideea was to substitute $t=frac{sin x}{2}, $ thus we have to evaluate $int_0^frac{pi}{2} sin x ln left(frac{sin x}{2}right) dx$ using for example beta function.
– Zacky
Nov 6 at 20:41
2
"derivatives of the Beat function are tricky, but they boil down to derivatives of the $psi$ function, which I am not very strong in" If you are not yet familiar with such functions, you should study it now. This question is really a straightforward application of beta function. They are indispensable if you want to go beyond Irresistible integrals.
– pisco
Nov 8 at 7:19
|
show 6 more comments
up vote
14
down vote
favorite
up vote
14
down vote
favorite
I have been trying to work out a question from Irresistible Integrals, but I don't know how to proceed. The exact problem is Exercise 13.7.1. It asks us to show that $$zeta(3)=4-zeta(2)+frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt$$
The $zeta$ in question is the Riemann Zeta Function. The book does this integral through something that its calls its "Master Formula," but asks the reader to check it directly, which I assume means to compute the integral directly. The book gives the hint of expanding the integrand as a series, using $$frac{1}{sqrt{1-4x}}=sum_{n=0}^inftybinom{2n}{n}x^n$$
After applying this, I have $$int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=sum_{n=0}^inftybinom{2n}{n}int_0^{1/2}t^{2n+1}ln^3(t)dt$$
$$=-sum_{n=0}^inftyfrac{1}{2^{2n+2}}binom{2n}{n}left(frac{ln^3(2)}{2n+2}+frac{3ln^2(2)}{(2n+2)^2}+frac{6ln(2)}{(2n+2)^3}+frac{6}{(2n+2)^4}right)$$
(If someone wants to check that to makes sure I'm right, then by all means, go for it) I get stuck from here. I have a few tricks to evaluate the sum for the first two terms, but not the last two terms. If you have any ideas on how to evaluate this sum (or an alternate way of evaluating this integral in the first place), let me know.
Thanks!
Edit: Considering that this integral is being multiplied by $8/3$ in the original problem, it may be convenient to fact or out the $8/3$ to get
$$frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=$$
$$-sum_{n=0}^inftyfrac{1}{3cdot2^{2n-1}}binom{2n}{n}left(frac{ln^3(2)}{2n+2}+frac{3ln^2(2)}{(2n+2)^2}+frac{6ln(2)}{(2n+2)^3}+frac{6}{(2n+2)^4}right)$$
It also might be more convenient to factor the $2$'s from the denominators and get
$$frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=$$
$$-sum_{n=0}^inftyfrac{1}{3cdot2^{2n}}binom{2n}{n}left(frac{ln^3(2)}{n+1}+frac{3}{2}frac{ln^2(2)}{(n+1)^2}+frac{3}{2}frac{ln(2)}{(n+1)^3}+frac{3}{4}frac{1}{(n+1)^4}right)$$
It looks a bit nicer, but I still don't know how to proceed.
Edit 2: A good suggestion is to turn this integral into a form of the Beta function. This can be done a few different ways, but if we take $u=4t^2$ we get
$$int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=frac{1}{8}int_0^1(1-u)^{-1/2}ln^3left(sqrt{frac{u}{4}}right)du$$
$$=frac{1}{64}int_0^1(1-u)^{-1/2}left(ln^3(u)-3ln^2(u)ln(4)+3ln(u)ln^24)-ln^3(4)right)du$$
$$=frac{1}{64}left(frac{partial^3B}{partial x^3}(1,1/2)-3ln(4)frac{partial^2B}{partial x^2}(1,1/2)+3ln^2(4)frac{partial B}{partial x}(1,1/2)-ln^3(4)B(1,1/2)right)$$
At this point, derivatives of the Beta function are tricky, but they boil down to derivatives of the $psi$ function, which I am not very strong in. If someone has a solution from here, that would be handy as well, but I am hoping that someone can give a solution somehow using a series approach, as suggested in the text.
calculus integration sequences-and-series
asked Nov 4 at 4:05
Josh B.
7098
I have been trying to work out a question from Irresistible Integrals, but I don't know how to proceed. The exact problem is Exercise 13.7.1. It asks us to show that $$zeta(3)=4-zeta(2)+frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt$$
The $zeta$ in question is the Riemann Zeta Function. The book does this integral through something that its calls its "Master Formula," but asks the reader to check it directly, which I assume means to compute the integral directly. The book gives the hint of expanding the integrand as a series, using $$frac{1}{sqrt{1-4x}}=sum_{n=0}^inftybinom{2n}{n}x^n$$
After applying this, I have $$int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=sum_{n=0}^inftybinom{2n}{n}int_0^{1/2}t^{2n+1}ln^3(t)dt$$
$$=-sum_{n=0}^inftyfrac{1}{2^{2n+2}}binom{2n}{n}left(frac{ln^3(2)}{2n+2}+frac{3ln^2(2)}{(2n+2)^2}+frac{6ln(2)}{(2n+2)^3}+frac{6}{(2n+2)^4}right)$$
(If someone wants to check that to makes sure I'm right, then by all means, go for it) I get stuck from here. I have a few tricks to evaluate the sum for the first two terms, but not the last two terms. If you have any ideas on how to evaluate this sum (or an alternate way of evaluating this integral in the first place), let me know.
Thanks!
Edit: Considering that this integral is being multiplied by $8/3$ in the original problem, it may be convenient to fact or out the $8/3$ to get
$$frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=$$
$$-sum_{n=0}^inftyfrac{1}{3cdot2^{2n-1}}binom{2n}{n}left(frac{ln^3(2)}{2n+2}+frac{3ln^2(2)}{(2n+2)^2}+frac{6ln(2)}{(2n+2)^3}+frac{6}{(2n+2)^4}right)$$
It also might be more convenient to factor the $2$'s from the denominators and get
$$frac{8}{3}int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=$$
$$-sum_{n=0}^inftyfrac{1}{3cdot2^{2n}}binom{2n}{n}left(frac{ln^3(2)}{n+1}+frac{3}{2}frac{ln^2(2)}{(n+1)^2}+frac{3}{2}frac{ln(2)}{(n+1)^3}+frac{3}{4}frac{1}{(n+1)^4}right)$$
It looks a bit nicer, but I still don't know how to proceed.
Edit 2: A good suggestion is to turn this integral into a form of the Beta function. This can be done a few different ways, but if we take $u=4t^2$ we get
$$int_0^{1/2}frac{tln^3(t)}{sqrt{1-4t^2}}dt=frac{1}{8}int_0^1(1-u)^{-1/2}ln^3left(sqrt{frac{u}{4}}right)du$$
$$=frac{1}{64}int_0^1(1-u)^{-1/2}left(ln^3(u)-3ln^2(u)ln(4)+3ln(u)ln^24)-ln^3(4)right)du$$
$$=frac{1}{64}left(frac{partial^3B}{partial x^3}(1,1/2)-3ln(4)frac{partial^2B}{partial x^2}(1,1/2)+3ln^2(4)frac{partial B}{partial x}(1,1/2)-ln^3(4)B(1,1/2)right)$$
At this point, derivatives of the Beta function are tricky, but they boil down to derivatives of the $psi$ function, which I am not very strong in. If someone has a solution from here, that would be handy as well, but I am hoping that someone can give a solution somehow using a series approach, as suggested in the text.
calculus integration sequences-and-series
calculus integration sequences-and-series
asked Nov 4 at 4:05
Josh B.
7098
asked Nov 4 at 4:05
Josh B.
7098
edited Dec 3 at 20:08
asked Nov 4 at 4:05
Josh B.
7098
asked Nov 4 at 4:05
Josh B.
7098
asked Nov 4 at 4:05
Josh B.
7098
7098
Have you tried evaluating a few iterations of n and seeing if there is a pattern in cancellation?
– Henry Lee
Nov 4 at 11:27
I have tried writing out a few terms to look for a pattern, but there wasn't anything obvious to me. If I'm interpreting you right, then I would say there is no cancellation, since everything inside the final sum is positive.
– Josh B.
Nov 4 at 16:09
Yes that sounds correct, it seems that the only way to do it would be to split it into several different summations and try to find a general formula for each one. But I am not sure that one will exist
– Henry Lee
Nov 4 at 16:11
1
My main ideea was to substitute $t=frac{sin x}{2}, $ thus we have to evaluate $int_0^frac{pi}{2} sin x ln left(frac{sin x}{2}right) dx$ using for example beta function.
– Zacky
Nov 6 at 20:41
2
"derivatives of the Beat function are tricky, but they boil down to derivatives of the $psi$ function, which I am not very strong in" If you are not yet familiar with such functions, you should study it now. This question is really a straightforward application of beta function. They are indispensable if you want to go beyond Irresistible integrals.
– pisco
Nov 8 at 7:19
|
show 6 more comments
Have you tried evaluating a few iterations of n and seeing if there is a pattern in cancellation?
– Henry Lee
Nov 4 at 11:27
I have tried writing out a few terms to look for a pattern, but there wasn't anything obvious to me. If I'm interpreting you right, then I would say there is no cancellation, since everything inside the final sum is positive.
– Josh B.
Nov 4 at 16:09
Yes that sounds correct, it seems that the only way to do it would be to split it into several different summations and try to find a general formula for each one. But I am not sure that one will exist
– Henry Lee
Nov 4 at 16:11
1
My main ideea was to substitute $t=frac{sin x}{2}, $ thus we have to evaluate $int_0^frac{pi}{2} sin x ln left(frac{sin x}{2}right) dx$ using for example beta function.
– Zacky
Nov 6 at 20:41
2
"derivatives of the Beat function are tricky, but they boil down to derivatives of the $psi$ function, which I am not very strong in" If you are not yet familiar with such functions, you should study it now. This question is really a straightforward application of beta function. They are indispensable if you want to go beyond Irresistible integrals.
– pisco
Nov 8 at 7:19
Have you tried evaluating a few iterations of n and seeing if there is a pattern in cancellation?
– Henry Lee
Nov 4 at 11:27
Have you tried evaluating a few iterations of n and seeing if there is a pattern in cancellation?
– Henry Lee
Nov 4 at 11:27
I have tried writing out a few terms to look for a pattern, but there wasn't anything obvious to me. If I'm interpreting you right, then I would say there is no cancellation, since everything inside the final sum is positive.
– Josh B.
Nov 4 at 16:09
I have tried writing out a few terms to look for a pattern, but there wasn't anything obvious to me. If I'm interpreting you right, then I would say there is no cancellation, since everything inside the final sum is positive.
– Josh B.
Nov 4 at 16:09
Yes that sounds correct, it seems that the only way to do it would be to split it into several different summations and try to find a general formula for each one. But I am not sure that one will exist
– Henry Lee
Nov 4 at 16:11
Yes that sounds correct, it seems that the only way to do it would be to split it into several different summations and try to find a general formula for each one. But I am not sure that one will exist
– Henry Lee
Nov 4 at 16:11
1
1
My main ideea was to substitute $t=frac{sin x}{2}, $ thus we have to evaluate $int_0^frac{pi}{2} sin x ln left(frac{sin x}{2}right) dx$ using for example beta function.
– Zacky
Nov 6 at 20:41
My main ideea was to substitute $t=frac{sin x}{2}, $ thus we have to evaluate $int_0^frac{pi}{2} sin x ln left(frac{sin x}{2}right) dx$ using for example beta function.
– Zacky
Nov 6 at 20:41
2
2
"derivatives of the Beat function are tricky, but they boil down to derivatives of the $psi$ function, which I am not very strong in" If you are not yet familiar with such functions, you should study it now. This question is really a straightforward application of beta function. They are indispensable if you want to go beyond Irresistible integrals.
– pisco
Nov 8 at 7:19
"derivatives of the Beat function are tricky, but they boil down to derivatives of the $psi$ function, which I am not very strong in" If you are not yet familiar with such functions, you should study it now. This question is really a straightforward application of beta function. They are indispensable if you want to go beyond Irresistible integrals.
– pisco
Nov 8 at 7:19
|
show 6 more comments
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Have you tried evaluating a few iterations of n and seeing if there is a pattern in cancellation?
– Henry Lee
Nov 4 at 11:27
I have tried writing out a few terms to look for a pattern, but there wasn't anything obvious to me. If I'm interpreting you right, then I would say there is no cancellation, since everything inside the final sum is positive.
– Josh B.
Nov 4 at 16:09
Yes that sounds correct, it seems that the only way to do it would be to split it into several different summations and try to find a general formula for each one. But I am not sure that one will exist
– Henry Lee
Nov 4 at 16:11
1
My main ideea was to substitute $t=frac{sin x}{2}, $ thus we have to evaluate $int_0^frac{pi}{2} sin x ln left(frac{sin x}{2}right) dx$ using for example beta function.
– Zacky
Nov 6 at 20:41
2
"derivatives of the Beat function are tricky, but they boil down to derivatives of the $psi$ function, which I am not very strong in" If you are not yet familiar with such functions, you should study it now. This question is really a straightforward application of beta function. They are indispensable if you want to go beyond Irresistible integrals.
– pisco
Nov 8 at 7:19