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Welcome to NRICH.

ò₀ ^¥[x³/(e^x-1)]dx

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By Arun Iyer on Wednesday, December 19, 2001 - 06:13 pm:

could anyone please evaluate....

ò₀ ^¥[x³/(e^x-1)]dx

love arun

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By Michael Doré on Thursday, December 20, 2001 - 01:04 am:

I would suggest splitting up x³/(e^x - 1) as:

x³e^-x/(1 - e^-x) = x³(e^-x + e^-2x + e^-3x + ...)

i.e. a geometric series (which we can do since e^-x < 1 for x > 0).

Note that ò₀ ^¥x³e^-nxdx = ò₀ ^¥u³e^-u/n ⁴ du = 6/n⁴.

So the answer is 6/1⁴ + 6/2⁴ + ... = 6z(4). I can't actually remember the value of z(4) offhand, but it is well known (it is a rational multiple of p⁴).

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By Arun Iyer on Thursday, December 20, 2001 - 05:58 pm:

Well, that's some neat transformation there...beautiful.

I know that z(4)=p⁴/90
However, do you know a way of proving this...Michael?

love arun

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By Arun Iyer on Saturday, December 22, 2001 - 06:42 pm:

Well, anyone know proof for z(4)=p⁴/90?

love arun

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By Yatir Halevi on Saturday, December 22, 2001 - 07:14 pm:

I know that it is not a proof
but i did see somewhere that:

z(2k)=2^2k-1p^2kB_k/(2k)!

k=1,2,3...

lets take k=2.
we get:
The second bernoulli number is 1/30
2³p⁴/30*24=p⁴/90

Yatir

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By Arun Iyer on Saturday, December 22, 2001 - 07:36 pm:

i know that formula...
i had given the link for that some months back..

this is the site..
http://numbers.computation.free.fr/Constants/Miscellaneous/bernoulli.html

love arun

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By Michael Doré on Sunday, December 23, 2001 - 01:44 am:

I think it is easy if you're willing to accept Euler's sine product. If we start with:

sin x = x(1 - x²/p²)(1 - x²/(4p²))...

then on taking logarithms and differentiating you get:

cot x = 1/x + S [1/(x + rp) + 1/(x - rp)]

where r is taken over the naturals.

Differentiate each side 2n-1 times:

d^2n-1/dx^2n-1cot x = -(2n-1)!(1/x²ⁿ + S 1/(x + rp)²ⁿ + S 1/(x - rp)²ⁿ)

Bring the 1/x²ⁿ term to the other side then take the limit as x->0:

lim(x->0)((d^2n-1/dx^2n-1cot x) + (2n-1)!/x²ⁿ) = -2(2n-1)!z(2n)/p²ⁿ

Write cot x = i(e^2ix + 1)/(e^2ix - 1), multiply through by -i and you get:

lim(x->0)(d^2n-1/dx^2n-1((e^2ix + 1)/(e^2ix - 1)) - i(2n-1)!/x²ⁿ) = 2i(2n-1)!z(2n)

So:

lim(x->0)(d^2n-1/dx^2n-1((e^2ix + 1)/(2ix) * 2ix/(e^2ix - 1)) - i(2n-1)!/x²ⁿ) = 2i(2n-1)!z(2n)

Setting y = 2ix and using the chain rule this becomes:

lim(y->0)(d^2n-1/dy^2n-1((e^y + 1)/y * y/(e^y - 1))*2^2n-1(-1)ⁿ⁺¹i + 2²ⁿ(-1)ⁿ⁺¹(2n-1)!i/y²ⁿ) = 2i(2n-1)!z(2n) (*)

Note that (e^y + 1)/y and y/(e^y - 1) can both be series expanded:

(e^y + 1)/y = 2/y + 1 + y/2! + y²/3! + y³/4!

y/(e^y - 1) = B₀ + B₁y + B₂y²/2! + ...

where B_n is the nth Bernoulli number (using the well fact that y/(e^y - 1) is the generating function for the Bernoulli numbers).

If you substitute these series in (*) and expand out the product then you should find that the reciprocal terms cancel (so luckily the function doesn't blow up at the origin) and then the final result follows on applying the (well known??) identity:

n!B_n = Sⁿ _r=1C(n,r)B_r

which is easily proved by equating coefficients in the identity:

x/(e^x - 1) * e^x = x + x/(e^x - 1)

(e^x+1)/(e^2x - 1) = 1/(e^x - 1).

Of course this can all be greatly speeded up for z(4).

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By Michael Doré on Sunday, December 23, 2001 - 02:56 am:

Oops, ignore the penultimate line. And I think the identity is only for n > 1.

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By Arun Iyer on Sunday, December 23, 2001 - 05:47 pm:

whoa!!pretty long proof....phew!!

thanks michael...

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