closed form for $sum_{ngeq1}pi nlogfrac{4n+1}{4n-1}prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$?
$begingroup$
Look at what I found!
$$int_{-pi/4}^{pi/4}xcsc x mathrm dx=sum_{ngeq1}pi nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
I want to know if it is valid, and if there is a nice closed form to go with my series.
Here's my proof.
Recall the famous product representation
$$sin x=xprod_{kgeq1}bigg(1-frac{x^2}{pi^2k^2}bigg)$$
Hence we have that
$$I=int_{-pi/4}^{pi/4}xcsc x mathrm dx=int_{-pi/4}^{pi/4}prod_{kgeq1}frac{pi^2k^2}{pi^2k^2-x^2}mathrm dx$$
We may now preform a fraction decomposition, and start with
$$prod_{kgeq1}frac{pi^2k^2}{pi^2k^2-x^2}=sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_n$$
Multiplying both sides by $prod_{kgeq1}frac{pi^2k^2-x^2}{pi^2k^2}$,
$$1=sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_nprod_{kgeq1}frac{pi^2k^2-x^2}{pi^2k^2}$$
$$1=sum_{ngeq1}b_nprod_{kgeq1\kneq n}frac{pi^2k^2-x^2}{pi^2k^2}$$
Which in turn implies that, for any $minBbb N$,
$$1=b_mprod_{kgeq1\kneq m}frac{pi^2k^2-pi^2m^2}{pi^2k^2}$$
$$b_m=prod_{kgeq1\kneq m}frac{k^2}{k^2-m^2}$$
Hence we can begin the process of integration:
$$I=int_{-pi/4}^{pi/4}sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_nmathrm dx$$
$$I=sum_{ngeq1}pi^2n^2b_nint_{-pi/4}^{pi/4}frac{mathrm dx}{pi^2n^2-x^2}$$
This final integral evaluates to
$$int_{-pi/4}^{pi/4}frac{mathrm dx}{pi^2n^2-x^2}=frac1{pi n}logbigg(frac{4n+1}{4n-1}bigg)$$
So finally we have
$$I=sum_{ngeq1}pi nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
Does this work? Is there a closed form value for $I$? Is there another way to prove this?
Update:
Integrating form $0$ to $pi/2$ instead of $-pi/4$ to $pi/4$ gives, according to Wolfram Alpha,
$$sum_{ngeq1}frac{pi n}2logbigg(frac{2n+1}{2n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}=2G$$
Where $G$ is Catalan's constant. Isn't that cool?! That's crazy cool!
integration closed-form alternative-proof
asked Dec 4 '18 at 18:34
clathratusclathratus
3,730333
$endgroup$
add a comment |
$begingroup$
Look at what I found!
$$int_{-pi/4}^{pi/4}xcsc x mathrm dx=sum_{ngeq1}pi nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
I want to know if it is valid, and if there is a nice closed form to go with my series.
Here's my proof.
Recall the famous product representation
$$sin x=xprod_{kgeq1}bigg(1-frac{x^2}{pi^2k^2}bigg)$$
Hence we have that
$$I=int_{-pi/4}^{pi/4}xcsc x mathrm dx=int_{-pi/4}^{pi/4}prod_{kgeq1}frac{pi^2k^2}{pi^2k^2-x^2}mathrm dx$$
We may now preform a fraction decomposition, and start with
$$prod_{kgeq1}frac{pi^2k^2}{pi^2k^2-x^2}=sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_n$$
Multiplying both sides by $prod_{kgeq1}frac{pi^2k^2-x^2}{pi^2k^2}$,
$$1=sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_nprod_{kgeq1}frac{pi^2k^2-x^2}{pi^2k^2}$$
$$1=sum_{ngeq1}b_nprod_{kgeq1\kneq n}frac{pi^2k^2-x^2}{pi^2k^2}$$
Which in turn implies that, for any $minBbb N$,
$$1=b_mprod_{kgeq1\kneq m}frac{pi^2k^2-pi^2m^2}{pi^2k^2}$$
$$b_m=prod_{kgeq1\kneq m}frac{k^2}{k^2-m^2}$$
Hence we can begin the process of integration:
$$I=int_{-pi/4}^{pi/4}sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_nmathrm dx$$
$$I=sum_{ngeq1}pi^2n^2b_nint_{-pi/4}^{pi/4}frac{mathrm dx}{pi^2n^2-x^2}$$
This final integral evaluates to
$$int_{-pi/4}^{pi/4}frac{mathrm dx}{pi^2n^2-x^2}=frac1{pi n}logbigg(frac{4n+1}{4n-1}bigg)$$
So finally we have
$$I=sum_{ngeq1}pi nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
Does this work? Is there a closed form value for $I$? Is there another way to prove this?
Update:
Integrating form $0$ to $pi/2$ instead of $-pi/4$ to $pi/4$ gives, according to Wolfram Alpha,
$$sum_{ngeq1}frac{pi n}2logbigg(frac{2n+1}{2n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}=2G$$
Where $G$ is Catalan's constant. Isn't that cool?! That's crazy cool!
integration closed-form alternative-proof
asked Dec 4 '18 at 18:34
clathratusclathratus
3,730333
$endgroup$
1
$begingroup$
Yes it is 'crazy-cool' ... & there are lirerally thousands of crazy-cool results like that! I can't get enough of 'em.
$endgroup$
– AmbretteOrrisey
Dec 4 '18 at 22:00
add a comment |
$begingroup$
Look at what I found!
$$int_{-pi/4}^{pi/4}xcsc x mathrm dx=sum_{ngeq1}pi nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
I want to know if it is valid, and if there is a nice closed form to go with my series.
Here's my proof.
Recall the famous product representation
$$sin x=xprod_{kgeq1}bigg(1-frac{x^2}{pi^2k^2}bigg)$$
Hence we have that
$$I=int_{-pi/4}^{pi/4}xcsc x mathrm dx=int_{-pi/4}^{pi/4}prod_{kgeq1}frac{pi^2k^2}{pi^2k^2-x^2}mathrm dx$$
We may now preform a fraction decomposition, and start with
$$prod_{kgeq1}frac{pi^2k^2}{pi^2k^2-x^2}=sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_n$$
Multiplying both sides by $prod_{kgeq1}frac{pi^2k^2-x^2}{pi^2k^2}$,
$$1=sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_nprod_{kgeq1}frac{pi^2k^2-x^2}{pi^2k^2}$$
$$1=sum_{ngeq1}b_nprod_{kgeq1\kneq n}frac{pi^2k^2-x^2}{pi^2k^2}$$
Which in turn implies that, for any $minBbb N$,
$$1=b_mprod_{kgeq1\kneq m}frac{pi^2k^2-pi^2m^2}{pi^2k^2}$$
$$b_m=prod_{kgeq1\kneq m}frac{k^2}{k^2-m^2}$$
Hence we can begin the process of integration:
$$I=int_{-pi/4}^{pi/4}sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_nmathrm dx$$
$$I=sum_{ngeq1}pi^2n^2b_nint_{-pi/4}^{pi/4}frac{mathrm dx}{pi^2n^2-x^2}$$
This final integral evaluates to
$$int_{-pi/4}^{pi/4}frac{mathrm dx}{pi^2n^2-x^2}=frac1{pi n}logbigg(frac{4n+1}{4n-1}bigg)$$
So finally we have
$$I=sum_{ngeq1}pi nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
Does this work? Is there a closed form value for $I$? Is there another way to prove this?
Update:
Integrating form $0$ to $pi/2$ instead of $-pi/4$ to $pi/4$ gives, according to Wolfram Alpha,
$$sum_{ngeq1}frac{pi n}2logbigg(frac{2n+1}{2n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}=2G$$
Where $G$ is Catalan's constant. Isn't that cool?! That's crazy cool!
integration closed-form alternative-proof
asked Dec 4 '18 at 18:34
clathratusclathratus
3,730333
$endgroup$
Look at what I found!
$$int_{-pi/4}^{pi/4}xcsc x mathrm dx=sum_{ngeq1}pi nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
I want to know if it is valid, and if there is a nice closed form to go with my series.
Here's my proof.
Recall the famous product representation
$$sin x=xprod_{kgeq1}bigg(1-frac{x^2}{pi^2k^2}bigg)$$
Hence we have that
$$I=int_{-pi/4}^{pi/4}xcsc x mathrm dx=int_{-pi/4}^{pi/4}prod_{kgeq1}frac{pi^2k^2}{pi^2k^2-x^2}mathrm dx$$
We may now preform a fraction decomposition, and start with
$$prod_{kgeq1}frac{pi^2k^2}{pi^2k^2-x^2}=sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_n$$
Multiplying both sides by $prod_{kgeq1}frac{pi^2k^2-x^2}{pi^2k^2}$,
$$1=sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_nprod_{kgeq1}frac{pi^2k^2-x^2}{pi^2k^2}$$
$$1=sum_{ngeq1}b_nprod_{kgeq1\kneq n}frac{pi^2k^2-x^2}{pi^2k^2}$$
Which in turn implies that, for any $minBbb N$,
$$1=b_mprod_{kgeq1\kneq m}frac{pi^2k^2-pi^2m^2}{pi^2k^2}$$
$$b_m=prod_{kgeq1\kneq m}frac{k^2}{k^2-m^2}$$
Hence we can begin the process of integration:
$$I=int_{-pi/4}^{pi/4}sum_{ngeq1}frac{pi^2n^2}{pi^2n^2-x^2}b_nmathrm dx$$
$$I=sum_{ngeq1}pi^2n^2b_nint_{-pi/4}^{pi/4}frac{mathrm dx}{pi^2n^2-x^2}$$
This final integral evaluates to
$$int_{-pi/4}^{pi/4}frac{mathrm dx}{pi^2n^2-x^2}=frac1{pi n}logbigg(frac{4n+1}{4n-1}bigg)$$
So finally we have
$$I=sum_{ngeq1}pi nlogbigg(frac{4n+1}{4n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}$$
Does this work? Is there a closed form value for $I$? Is there another way to prove this?
Update:
Integrating form $0$ to $pi/2$ instead of $-pi/4$ to $pi/4$ gives, according to Wolfram Alpha,
$$sum_{ngeq1}frac{pi n}2logbigg(frac{2n+1}{2n-1}bigg)prod_{kgeq1\kneq n}frac{k^2}{k^2-n^2}=2G$$
Where $G$ is Catalan's constant. Isn't that cool?! That's crazy cool!
integration closed-form alternative-proof
integration closed-form alternative-proof
asked Dec 4 '18 at 18:34
clathratusclathratus
3,730333
asked Dec 4 '18 at 18:34
clathratusclathratus
3,730333
edited Dec 4 '18 at 21:55
clathratus
asked Dec 4 '18 at 18:34
clathratusclathratus
3,730333
asked Dec 4 '18 at 18:34
clathratusclathratus
3,730333
asked Dec 4 '18 at 18:34
clathratusclathratus
3,730333
3,730333
1
$begingroup$
Yes it is 'crazy-cool' ... & there are lirerally thousands of crazy-cool results like that! I can't get enough of 'em.
$endgroup$
– AmbretteOrrisey
Dec 4 '18 at 22:00
add a comment |
1
$begingroup$
Yes it is 'crazy-cool' ... & there are lirerally thousands of crazy-cool results like that! I can't get enough of 'em.
$endgroup$
– AmbretteOrrisey
Dec 4 '18 at 22:00
1
1
$begingroup$
Yes it is 'crazy-cool' ... & there are lirerally thousands of crazy-cool results like that! I can't get enough of 'em.
$endgroup$
– AmbretteOrrisey
Dec 4 '18 at 22:00
$begingroup$
Yes it is 'crazy-cool' ... & there are lirerally thousands of crazy-cool results like that! I can't get enough of 'em.
$endgroup$
– AmbretteOrrisey
Dec 4 '18 at 22:00
add a comment |
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$begingroup$
Yes it is 'crazy-cool' ... & there are lirerally thousands of crazy-cool results like that! I can't get enough of 'em.
$endgroup$
– AmbretteOrrisey
Dec 4 '18 at 22:00