# Prove the zeta log gamma integral $\int_0^1{\zeta(1-n,1-t)\ln\Gamma{(t)}}dt = \frac{H_n\zeta(-n)+\zeta’(-n)}{n}$

How to prove that $$\displaystyle\int_0^1{\zeta(1-n,1-t)\ln\Gamma{(t)}}dt = \frac{H_n\zeta(-n)+\zeta’(-n)}{n}$$?

For integer values Wolfram Alpha gave me solutions to the integral in the form on the right hand side, so I guessed the form.

It appears to break down when $$n$$ is less than around $$-\frac12$$. $$\zeta(s,a)$$ and $$\zeta(s)$$ are the Hurwitz zeta and Riemann zeta functions. $$H_n$$ are the harmonic numbers.

How can we prove it?

In order to obtain the closed form for integer $$n$$, in Wolfram Alpha, we can first rewrite it in terms of the Polygamma function.

First change the zeta function to Bernoulli Polynomials for easier manipulation.

\begin{align} &\frac{1}{\left(n-1\right)!}\int_{0}^{1}\zeta\left(1-n,1-t\right)\ln\Gamma\left(t\right)dt \\=& \ -\frac{1}{n!}\int_{0}^{1}B_n\left(1-t\right)\ln\Gamma\left(t\right)dt \\=& \ \frac{1}{n!}\int_{0}^{1}n\left(1-t\right)^{n-1}\ln\Gamma\left(t\right)dt-\frac{1}{n!}\int_{0}^{1}B_n\left(2-t\right)\ln\Gamma\left(t\right)dt \end{align}

This result can actually be generalized a bit by introducing the variable $$a$$.

\begin{align} &\frac{1}{n!}\int_{0}^{a}n\left(a-t\right)^{n-1}\ln\Gamma\left(t\right)dt-\frac{1}{n!}\int_{0}^{1}B_n\left(a+1-t\right)\ln\Gamma\left(t\right)dt \\=& \ \psi^{(-n-1)}\left(a\right)-\frac{1}{n!}\int_{0}^{1}B_n\left(a+1-t\right)\ln\Gamma\left(t\right)dt \end{align}

where $$\psi^{(s)}(a)$$ is the polygamma function. Here we are using this definition for negative order. Next we can utilize the definition of the bernoulli polynomials to obtain.

\begin{align} &\psi^{(-n-1)}\left(a\right)-\frac{1}{n!}\int_{0}^{1}\sum_{k=0}^{n}\binom{n}{k}B_k\left(a\right)\left(1-t\right)^{n-k}\ln\Gamma\left(t\right)dt \\=& \ \psi^{(-n-1)}\left(a\right)-\sum_{k=0}^{n}\frac{B_k\left(a\right)}{k!}\left(\frac{1}{\left(n-k\right)!}\int_{0}^{1}\left(1-t\right)^{n-k}\ln\Gamma\left(t\right)dt\right) \\=& \ \psi^{(-n-1)}\left(a\right)-\sum_{k=0}^{n}\frac{B_k\left(a\right)}{k!}\psi^{(-n-2+k)}\left(1\right) \end{align}

This last form is useful for plugging into Wolfram Alpha and generating our desired form. After analyzing patterns for several integers we see it equals

\begin{align} &-H_n\frac{B_{n+1}\left(a\right)}{\left(n+1\right)!}+\frac{\zeta'\left(-n\right)}{n!}+\frac{1}{n!}\sum_{k=1}^{a-1}k^{n}\ln k \\=& \ \frac{H_n\zeta\left(-n,a\right)}{n!}+\frac{\zeta'\left(-n,a\right)}{n!} \end{align}

Multiplying by $$(n-1)!$$ and setting $$a=1$$ yields the final result.

• Where did you get this question from ?
– pie
Dec 27, 2023 at 20:40
• @pie I invented this problem. The integral arose from work on the integration of the generating function of the Bernoulli polynomials divided by a power term. I intend to add some additional details for how I got WA to provide the closed form. Dec 28, 2023 at 19:57
• Thanks for responding. I thought this question came from a book and I wanted to know if that was true what is this book.
– pie
Dec 28, 2023 at 20:00

The equality (under the usual convention $$H_s:=\gamma+\psi(1+s)$$ for complex $$s$$) $$\mathscr{L}(s):=\int_0^1\zeta(1-s,1-t)\ln\Gamma(t)\,dt=\frac1s\big(H_s\zeta(-s)+\zeta'(-s)\big)$$ for, say, $$\Re s>1$$ can be viewed as an application of Parseval's theorem to the series $$\color{blue}{\ln\Gamma(t)=\frac{\ln2\pi}2+\sum_{n=1}^\infty\left(\frac1{2n}\cos 2n\pi t+\frac{\gamma+\ln2n\pi}{n\pi}\sin 2n\pi t\right)}$$ for $$0 (a variation of the known expansion; the formula $$\ln\Gamma(1-z)=\int_0^\infty\left(\frac{e^{zt}-1}{e^t-1}-ze^{-t}\right)\frac{dt}{t}$$ is a possible way to obtain this expansion) and the series (for $$\Re s>1$$ said) $$\color{blue}{\zeta(1-s,1-t)=\frac{2\Gamma(s)}{(2\pi)^s}\sum_{n=1}^\infty\frac1{n^s}\cos\left(\frac{\pi s}2+2n\pi t\right)}$$ obtained (e.g.) as follows. The definition $$\zeta(s,a)=\sum_{n=0}^\infty(n+a)^{-s}$$ leads to $$\Gamma(s)\zeta(s,a)=\int_0^\infty\frac{t^{s-1}e^{-at}}{1-e^{-t}}\,dt\qquad(a>0,\Re s>1)$$ which can be used to obtain (another known integral representation) $$\frac1{2\pi i}\int_\lambda\frac{z^{s-1}e^{az}}{1-e^z}\,dz=\frac{\zeta(s,a)}{\Gamma(1-s)},$$ where the contour $$\lambda$$ encircles the negative real axis (but not the poles $$z=2n\pi i$$, $$n\in\mathbb{Z}_{\neq0}$$ of the integrand). This is analytically continued (w.r.t. $$s$$) and, for $$\Re s<0$$ and $$0, the integral can be evaluated using residues (and considering, as a contour, a big rectangle with a notch around the negative real axis). This leads to the expansion of $$\zeta(1-s,1-t)$$ shown above.

Thus, Parseval's theorem yields $$\mathscr{L}(s)=\frac{\Gamma(s)}{(2\pi)^s}\left[\left(\frac12\cos\frac{\pi s}2-\frac{\gamma+\ln2\pi}\pi\sin\frac{\pi s}2\right)\zeta(1+s)+\frac1\pi\sin\frac{\pi s}2\zeta'(1+s)\right],$$ which simplifies to the announced result, using Riemann's functional equation for $$\zeta(1+s)$$.

Of course, the result holds analytically continued (from $$\Re s>1$$ to ???).

• Great solution! I too wonder for which complex $s$ it holds analytically continued. Jan 2 at 16:51

Congratulations on discovering this beautiful formula.

I will try to prove it for $$n$$ is a natural number, and this paper will help you to get it for all $$n\in\mathbb{C}$$.

First, let’s use King’s Rule, which is $$\int_0^1 f(t) dt=\int_0^1 f(1-t) dt$$ So the integral can be defined as (with denote the integral by $$\Omega_n$$) $$\Omega_n=\int_0^1 \zeta(1-n,1-t) \ln \Gamma(t) dt=\int_0^1 \zeta(1-n,t) \ln \Gamma(1-t) dt$$ then we get $$\big($$by using the fact $$x=y=\frac{x+y}{2} \big)$$ $$\int_0^1 \zeta(1-n,1-t) \ln \Gamma(t) dt=\frac{1}{2}\int_0^1 \left(\zeta(1-n,1-t) \ln \Gamma(t)+\zeta(1-n,t) \ln \Gamma(1-t) \right) dt$$ But we have for $$n$$ a natural number $$\zeta(1-n,1-t)=(-1)^n\zeta(1-n,t)$$ also for $$t \in (0,1)$$ $$\Gamma(1-t)=\frac{\pi}{\Gamma(t) \sin(\pi t)}$$ So $$\Omega_n=\frac{1}{2}\int_0^1 \zeta(1-n,1-t) \left( \ln \Gamma(t)+(-1)^n \ln \frac{\pi}{\Gamma(t) \sin(\pi t)} \right) dt$$ now its easy to show that $$\int_0^1 \zeta(1-n,1-t) dt=0$$ So then odd and even orders will defined $$\Omega_{2n}=-\frac{1}{2}\int_0^1 \zeta(1-2n,1-t) \ln \sin(\pi t) dt$$ $$\Omega_{2n+1}=\int_0^1 \zeta(-2n,1-t) \left(\ln \Gamma(t)+\frac{1}{2}\ln \sin(\pi t)\right) dt$$ now lets use this three series for $$t \in (0,1)$$

$$\displaystyle \zeta(1-n,1-t)=2(n-1)!\sum_{k=1}^{\infty}\frac{\cos\left(\frac{\pi n}{2}+2\pi kt\right)}{(2\pi k)^n}$$

$$\displaystyle \ln\sin(\pi t)=-\ln2-\sum_{k=1}^\infty \frac{\cos(2\pi kt)}{k}$$

$$\displaystyle \ln \Gamma(t)+\frac{1}{2} \ln \sin(\pi t)=\left(\frac{1}{2}-t \right)(\gamma+\ln2)+(1-t)\ln\pi+\frac{1}{\pi} \sum_{k=1}^{\infty} \frac{\ln k}{k} \sin(2\pi kt)$$

as we wrote that integral of the zeta function is zero here so all constants (in the second and third series I put) will vanished and then we have $$\Omega_{2n}=(2n-1)!\sum_{k=1}^{\infty}\frac{1}{(2\pi k)^{2n}}\sum_{p=1}^\infty \frac{1}{p} (-1)^n\int_0^1 \cos\left(2\pi kt\right) \cos(2\pi pt) dt$$ now it's easy to show that $$\int_0^1 \cos\left(2k\pi t\right) \cos(2\pi pt) dt=\frac{1}{2}, k=p$$ where its zero if $$k\ne p$$ therefore by putting $$p=k$$ $$\Omega_{2n}=(2n-1)!(-1)^n\sum_{k=1}^{\infty}\frac{1}{(2\pi k)^{2n}}\frac{1}{k} \frac{1}{2} =\frac{(2n-1)!(-1)^n}{2(2\pi)^{2n}}\zeta(2n+1)$$ and for odd orders we have (also by vanish the constants from ln gamma series) $$\Omega_{2n+1}=\int_0^1 \zeta(-2n,1-t) \left(-t(\gamma+\ln2+\ln\pi)+\frac{1}{\pi} \sum_{p=1}^{\infty} \frac{\ln p}{p} \sin(2\pi pt)\right) dt$$ using the series for zeta function and the value $$\int_0^1 \zeta(-2n,1-t) t dt=-\frac{\zeta(-2n-1)}{2n+1}$$ So $$\Omega_{2n+1}=\frac{\zeta(-2n-1)}{2n+1} (\gamma+\ln(2\pi))+\frac{2(2n)!}{\pi} \sum_{k=1}^{\infty}\frac{1}{(2\pi k)^{2n+1}} \sum_{p=1}^{\infty} \frac{\ln p}{p}(-1)^{n+1}\int_0^1 \sin\left(2\pi kt\right) \sin(2\pi pt) dt$$ now we its easy to evaluate (where $$n$$ and $$p$$ and $$k$$ are natural numbers) $$\int_0^1 \sin\left(2\pi kt\right) \sin(2\pi pt) dt=\frac{1}{2} ,k=p$$ where its zero for $$k\ne p$$ So we get the series $$\Omega_{2n+1}=\frac{\zeta(-2n-1)(-1)^n}{2n+1} (\gamma+\ln(2\pi))-\frac{2(2n)!(-1)^{n}}{\pi} \sum_{k=1}^{\infty}\frac{1}{(2\pi k)^{2n+1}} \frac{\ln k}{k} \frac{1}{2}$$ finally $$\Omega_{2n+1}=\frac{\zeta(-2n-1)}{2n+1} (\gamma+\ln(2\pi))+\frac{(2n)!(-1)^n}{\pi (2\pi)^{2n+1}} \zeta'(2n+2)$$ now by using the zeta function identity $$\zeta(2n+1)=\frac{(-1)^n(2\pi)^{2n+1}}{\pi(2n)!}\zeta'(-2n)$$ So we get $$\Omega_{2n}=\frac{\zeta'(-2n)}{2n}$$ also $$\zeta'(2n+2)=\frac{(-1)^n(2\pi)^{2n+2}}{2(2n+1)!} \left(\zeta'(-2n-1)+(H_{2n+1}-\ln(2\pi)-\gamma) \zeta(-2n-1)\right)$$ So $$\Omega_{2n+1}=\frac{H_{2n+1}\zeta(-2n-1)+\zeta'(-2n-1)}{2n+1}$$ So from $$\Omega_{2n},\Omega_{2n+1}$$ and since $$\zeta(-2n)=0$$ we get $$\Omega_n=\int_0^1 \zeta(1-n,1-t) \ln \Gamma(t) dt=\frac{H_{n}\zeta(-n)+\zeta'(-n)}{n}$$