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I was reading Irresistible Integrals by Victor H. Moll, where I encountered the following Taylor series expansion of $\ln\Gamma(1+x)$ $$ \ln\Gamma(1+x)=-\gamma x + \sum_{k=2}^{\infty}\dfrac{(-1)^k \zeta (k)}{k}x^k. $$

I am looking for the Taylor series for $\ln^2\Gamma(1+x)$. I guess that maybe I can just square the result I already have, like this $$ \begin{align} \ln^2\Gamma(1+x) &= \left(-\gamma x +\sum_{k=2}^{\infty}\dfrac{(-1)^k \zeta (k)}{k}x^k\right)^2\\ &= \gamma^2x^2 -2\gamma \sum_{k=2}^\infty \dfrac{(-1)^k\zeta(k)x^{k+1}}{k}+ \left(\sum_{k=2}^{\infty}\dfrac{(-1)^k \zeta (k)}{k}x^k\right)^2. \end{align} $$

If this is true, how can I square the sum? I think Cauchy product can help, but that will be very complicated and requires more manipulations which are way too much for me as I am new to the manipulations of zeta function.

Any help?

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  • $\begingroup$ $(\sum_{n=a}^b a_n)^2= \sum_ {k=a}^b a_k\sum_{n=a}^b a_n= \sum_{n=a}^b \sum_{k=a}^b a_n a_k$, but this is just basic factoring. Do you know about Cauchy Products? $\endgroup$ Commented Jun 21, 2021 at 18:31
  • $\begingroup$ Use en.wikipedia.org/wiki/… $\endgroup$
    – Gary
    Commented Jun 21, 2021 at 18:38

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$$S=\left(\sum_{n=a}^b a_n\right)^2= \sum_ {k=a}^b a_k\sum_{n=a}^b a_n= \sum_{n=a}^b \sum_{k=a}^b a_n a_k$$ but this is just basic factoring. Do you know about Cauchy Products?

I will work on this a bit more. Simply use the formula. This also works for other powers.

$$S= \left(\sum_{k=2}^{\infty}\dfrac{(-1)^k \zeta (k)}{k}x^k\right)^2= \sum_{k=2}^{\infty}\dfrac{(-1)^k \zeta (k)}{k}x^k \sum_{n=2}^{\infty}\dfrac{(-1)^n \zeta (n)}{n}x^n= \sum_{k=2}^{\infty} \sum_{n=2}^{\infty}\dfrac{(-1)^k \zeta (k)}{k}x^k\dfrac{(-1)^n \zeta (n)}{n}x^n= \sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \dfrac{(-1)^{k+n} \zeta(n)\zeta (k)}{kn}x^{k+n} $$

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  • $\begingroup$ I had thought of reindexing. Thanks. $\endgroup$ Commented Jun 21, 2021 at 18:35

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