OMG !! Zeta and Pi in one integral!

Calculus Level 5

If $\int_0^1 \frac{1-x}{1-x^6}(\ln x)^4 \ \mathrm{d}x= \frac{a\pi^k}{b\sqrt{c}} + \frac{d\zeta(5)}{e}$ Where $a,b,c,d,e,k$ are positive integers and $c$ is not divisible by any perfect square. Find $a+b+c+d+e+k$

Details and assumptions : $\zeta(5) = \sum_{k=1}^{\infty} \frac{1}{k^5} .$

The answer is 926.

2 solutions

Pranav Arora
Jul 17, 2014

Consider the integral:

$I(a)=\int_0^1 x^a\frac{1-x}{1-x^6}\,dx$

To evaluate the above, use the series representation of $\dfrac{1}{1-x^6}$ i.e

$I(a)=\sum_{k=0}^{\infty} \int_0^1 x^a(1-x)x^{6k}=\sum_{k=0}^{\infty} \left(\frac{1}{6k+a+1}-\frac{1}{6k+a+2}\right)$

Differentiate $I$ wrt $a$ four times and substitution $a=0$ , hence,

$\begin{aligned} \int_0^1 \frac{1-x}{1-x^6}(\ln x)^4\,dx &=24\sum_{k=0}^{\infty}\left(\frac{1}{(6k+1)^5}-\frac{1}{(6k+2)^5}\right) \\ &=\frac{24}{6^5}\sum_{k=0}^{\infty}\left(\frac{1}{(k+1/6)^5}-\frac{1}{(k+1/3)^5}\right) \\ &=\frac{24}{6^5}\left(\frac{-\psi_4(1/6)}{24}-\frac{-\psi_4(1/3)}{24}\right) \\ &=\frac{1}{6^5}\left(\psi_4(1/3)-\psi_4(1/6)\right) \\ &=\boxed{\dfrac{605 \zeta(5)}{54}+\dfrac{16\pi^5}{243\sqrt{3}}} \\ \end{aligned}$

$\psi_n(x)$ is the $n^{th}$ derivative of digamma function.

Nice solution Pranav! I reached up to this stage:

$24\displaystyle\sum_{k=0}^{\infty}\bigg( \frac{1}{(6k+1)^{5}}-\frac{1}{(6k+2)^{5}}\bigg)$

but as I really don't know anything about the digamma function I could not proceed further. Could you please suggest me some resources to learn about functions such as gamma function, beta function etc. from the basics. I have seen that it is used in many tough integration problems. Thanks! $\ddot\smile$

Karthik Kannan - 6 years, 11 months ago

To be honest, I never properly learned them from any book. The only resources I have used are the wiki and mathworld pages and some solutions to problems involving such functions. You can look over the internet for some resources. Also, beta and gamma functions might be in your syllabus this year. I still need to learn more about digamma functions, its been only a few days I came to know about them.

Pranav Arora - 6 years, 11 months ago

Thanks for the help Pranav! $\ddot\smile$

Karthik Kannan - 6 years, 10 months ago

@Sudeep Salgia : If you don't mind, can you please share your solution to this problem? My proof for the final equality is quite long and I don't like it. Do you have some other method? Thanks!

Pranav Arora - 6 years, 11 months ago

Well I faced the same problem as @Karthik Kannan did and had to use W|A to evaluate the sum :P.
I had tried when I solved the sum but did not get any method for the same but I will give it a shot once again maybe with some knowledge of digamma functions etc.

Sudeep Salgia - 6 years, 11 months ago

@Pranav Arora Cool, but the latter identity seems to be tricky.

Have you noticed that the fraction has a partial fraction decomposition that might help?

Haroun Meghaichi - 6 years, 11 months ago

I am not sure but are you talking about this:

$\frac{1-x}{1-x^6}=\frac{1-x}{(1-x^3)(1+x^3)}=\frac{(1-x)}{2}\left(\frac{1}{1-x^3}+\frac{1}{1+x^3}\right)$

Pranav Arora - 6 years, 11 months ago

No, it can be decomposed more than this.

Haroun Meghaichi - 6 years, 11 months ago

Did the exact same! Nice problem!

Kartik Sharma - 5 years, 10 months ago

Haroun Meghaichi
Aug 12, 2014

@Pranav Arora Here's my solution . partial decomposition also get you to $\psi_4(1/3)$ but it is longer.

OMG !! Zeta and Pi in one integral!

The answer is 926.

2 solutions

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