complex-analysisdigamma-functiongamma functionintegrationreference-request
I was reading about the Digamma function, defined as:
$$\psi(z) = \frac{d}{dx}\ln( \Gamma(z)) = \frac{\Gamma ' (z)}{\Gamma(z)}$$
And the following integral representation of $\psi(z)$ was given for $z:\Re(z) > 0$:
$$\psi(z) = \int_0^{\infty}\Big{(}\frac{e^{-t}}{t} – \frac{e^{-zt}}{1-e^{-t}}\Big{)}dt$$
The proof of which is written off as "due to Gauss," but I cannot find his proof. What was Gauss's proof of this?
Hints/guidelines:
$1$. Prove that $\Gamma(z)$ has no zeroes so $\psi(z)$ can have no singular points other than the poles $0,-1,-2,...$. Then it's not too hard to prove using the properties of $\Gamma$ function that $\psi$ has the representation $$ \psi(z)=-\frac{1}{z+n}+\mathcal{E}(z+n) $$ Here $\mathcal{E}(z+n)$ is the regular part of $\psi(z)$.
$2$. See the comment.
$3.$ Non-detailed proof:
Using the definition $\Gamma'(z)$ and by replacing the $\text{log}(t)$ term with the integral $$ \text{log}(t)= \int_0^{\infty}\frac{e^{-x}-e^{-xt}}{x}, dx $$ we can show that $$ \Gamma'(z)=\int_0^\infty \frac{dx}{x}\left[ e^{-x}\Gamma(z)-\int_0^\infty e^{-t(x+1)}t^{z-1} dt \right]. $$ Using the substitution $u=t(x+1)$ we get $$ \psi(z)=\int_0^\infty\left[e^{-x}-\frac{1}{(x+1)^{z}} \right]\frac{dx}{x}. $$ Using this we can find $$ \psi(z)=\lim_{\epsilon \to 0} \left[ \int_\epsilon^\infty \frac{e^{-x}}{x}dx-\int_\epsilon^\infty \frac{1}{(x+1)^{z}x} dx \right]. $$ Substituting $1+x = e^u$ for the last integral we get $$ \psi(z)=\lim_{\epsilon \to 0} \left[ \int_{log(1+\epsilon)}^\infty \left( \frac{e^{-u}}{u}-\frac{e^{-uz}}{1-e^{-u}}\right)du -\int_{log(1+\epsilon)}^\epsilon \frac{e^{-u}}{u} du \right]. $$ Here the last integral goes to $0$ as $\epsilon \to 0$ so we get $$ \psi(z)= \int_{0}^\infty \left( \frac{e^{-u}}{u}-\frac{e^{-uz}}{1-e^{-u}}\right)du. $$ Plugging in $z=1$, subtracting the result and using substitution $x=e^{-u}$ we finally get $$ \psi(z)=-\gamma+\int_0^{1}\frac{1-x^{z-1}}{1-x}dx. $$ Now the claim follows by using $(1-x)^{-1}=\sum_{n=0}^{\infty} x^n$ and integrating term by term.
As I mentioned in my question, when integrating back from $0$ to $z$ the L.H.S. is not our goal integral, nevertheless, we can still evaluate the integral by this method. Here is how to do so:
First, lets integrate back the first integral, namely:
$$ \begin{aligned} &\int_0^z\,\int_0^\infty\frac{2wx }{(w^2+x^2)(e^{2\pi x}-1)}\,dx\,dw=\int_0^zw\log(w)\,dw-\int_0^zw\psi(w)\,dw-\frac12\int_0^zdw\\ &\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx-2\int_0^\infty\frac{x \ln(x)}{e^{2\pi x}-1}\,dx=\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\ &\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx=2\int_0^\infty\frac{x \ln(x)}{e^{2\pi x}-1}\,dx+\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\ &\int_0^\infty\frac{x \ln(z^2+x^2)}{e^{2\pi x}-1}\,dx=\zeta^{\prime}(-1)+\frac{z^2 \ln z}{2}-\frac{z^2}{4}-z \ln \Gamma(z)+\int_0^z\ln \Gamma(w)dw-\frac{z}{2}\\ \end{aligned} $$
Taking the limit $z \to 1$
$$\int_0^\infty\frac{x \ln(1+x^2)}{e^{2\pi x}-1}\,dx=\zeta^{\prime}(-1)-\frac{3}{4}+\frac12 \ln 2 \pi \qquad \blacksquare$$
Integrating back the second integral:
$$\begin{aligned} &\int_0^z\,\int_0^\infty\frac{2wx }{(w^2+x^2)(e^{\pi x}-1)}\,dx\,dw=\int_0^zw\log\left(\frac{w}{2}\right)\,dw-\int_0^z w\psi\left(\frac{w}{2}\right)\,dw-\int_0^z dw\\ &\int_0^\infty\frac{x \ln(z^2+x^2) }{(e^{\pi x}-1)}\,dx=2\int_0^\infty\frac{x \ln(x) }{(e^{\pi x}-1)}\,dx+\frac{z^2\ln z}{2}-\frac{z^2}{4}-\frac{z^2 \ln 2 }{2}-z-4\int_0^{z/2} w\psi\left(w\right)\,dw\\ &\int_0^\infty\frac{x \ln(z^2+x^2) }{(e^{\pi x}-1)}\,dx=2\int_0^\infty\frac{x \ln(x) }{(e^{\pi x}-1)}\,dx+\frac{z^2\ln z}{2}-\frac{z^2}{4}-\frac{z^2 \ln 2 }{2}-z-4\left(\frac{z}{2} \ln\Gamma\left(\frac{z}{2}\right) -\int_0^{z/2} \ln\Gamma\left(w\right)\,dw \right)\\ \end{aligned} $$
Letting $z \to 1$
$$ \begin{aligned} \int_0^\infty\frac{x \ln(1+x^2) }{(e^{\pi x}-1)}\,dx&=2\left(\frac16+\frac{\ln 2}{6}-2 \ln A \right)-\frac{1}{4}-\frac{ \ln 2 }{2}-1-4\left(\frac{1}{2} \ln\Gamma\left(\frac{1}{2}\right) -\int_0^{1/2} \ln\Gamma\left(w\right)\,dw \right)\\ &=\frac13-4 \ln A -\frac{5}{4}-\frac{ \ln 2 }{6}-4\left(\frac{1}{4} \ln \pi -\left( \frac32\ln A+\frac{5}{24}\ln 2+\frac14 \ln \pi\right)\right)\\ &=\frac13-4 \ln A -\frac{5}{4}-\frac{ \ln 2 }{6}+6\ln A+\frac56\ln 2\\ &=2\ln A -\frac{11}{12}+\frac{ 2 }{3}\ln 2\\ &=-2\zeta^{\prime}(-1)+\frac{ 2 }{3}\ln 2-\frac34 \qquad \blacksquare\\ \end{aligned} $$
We the conclude
$$ \begin{aligned} &\int_{0}^{\infty} \frac{x \ln \left(1+x^{2}\right)}{e^{2 \pi x}-1} d x=\zeta^{\prime}(-1)+\ln \sqrt{2 \pi}-\frac{3}{4} \\ & \\ &\int_{0}^{\infty} \frac{x \ln \left(1+x^{2}\right)}{e^{\pi x}-1} d x=-2 \zeta^{\prime}(-1)+\frac{2}{3} \ln 2-\frac{3}{4} \end{aligned} $$
Therefore our integral equals
$$ \begin{aligned} \int_0^\infty\frac{x \ln(1+x^2)}{\sinh \pi x}\,dx&=2\left(-2 \zeta^{\prime}(-1)+\frac{2}{3} \ln 2-\frac{3}{4}-\zeta^{\prime}(-1)-\ln \sqrt{2 \pi}+\frac{3}{4}\right)\\ &=2\left(-3 \zeta^{\prime}(-1)+\frac{2}{3}\ln 2-\frac12 \ln 2-\frac12 \ln \pi\right)\\ &=-6 \zeta^{\prime}(-1)+\frac{4}{3}\ln 2- \ln 2- \ln \pi\\ &=-6 \zeta^{\prime}(-1)+\frac{\ln 2}{3}- \ln \pi\\ &=-6 \left(\frac{1}{12}-\ln A \right)+\frac{\ln 2}{3}- \ln \pi\\ &=-\frac{\ln e}{2}+\ln A^6 +\frac{\ln 2}{3}- \ln \pi\\ &=\ln \, \frac{\sqrt[3]{2}\, A^6}{\pi \sqrt{e}} \qquad \blacksquare \end{aligned} $$
Appendix
Consider the Integral
$$ \begin{aligned} I(s)&=\int_0^\infty \frac{x^{s-1}}{e^{cx}-1}\,dx\\ &=\frac{1}{c^s}\int_0^\infty \frac{x^{s-1}}{e^{x}-1}\,dx\\ &=\frac{1}{c^s}\int_0^\infty \frac{e^{-x}x^{s-1}}{1-e^{-x}}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \, \int_0^\infty x^{s-1}e^{-kx}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \frac{1}{k^s} \, \int_0^\infty x^{s-1}e^{-x}\,dx\\ &=\frac{1}{c^s}\sum_{k=1}^\infty \frac{1}{k^s} \, \int_0^\infty x^{s-1}e^{-x}\,dx\\ &=\frac{\Gamma(s)\zeta(s)}{c^s} \end{aligned} $$
Differentiating both sides w.r. to s
$$ \begin{aligned} \int_0^\infty \frac{x^{s-1} \ln x}{e^{cx}-1}\,dx&=\frac{d}{ds}\left(\frac{\Gamma(s)\zeta(s)}{c^s} \right)\\ &=\frac{c^s\left(\Gamma^{\prime}(s)\zeta(s)+\Gamma(s)\zeta^{\prime}(s) \right)-c^s\ln c \left( \Gamma(s)\zeta(s) \right)}{(c^s)^2}\\ &=\frac{\left(\Gamma(s)\psi(s)\zeta(s)+\Gamma(s)\zeta^{\prime}(s) \right)-\ln c \left( \Gamma(s)\zeta(s) \right)}{c^s}\\ \end{aligned} $$
Setting $s=2$
$$ \begin{aligned} \int_0^\infty \frac{x \ln x}{e^{cx}-1}\,dx&=\frac{\left(\Gamma(2)\psi(2)\zeta(2)+\Gamma(2)\zeta^{\prime}(2) \right)-\ln c \left( \Gamma(2)\zeta(2) \right)}{c^2}\\ &=\frac{\left((1-\gamma)\zeta(2)+\zeta(2)\left(\ln 2\pi+\gamma-12 \ln A\right) \right)-\zeta(2)\ln c }{c^2}\\ &=\zeta(2)\left(\frac{1-\ln c+\ln 2\pi-12 \ln A }{c^2}\right)\\ \end{aligned} $$
Setting $c=\pi$
$$\int_0^\infty \frac{x \ln x}{e^{\pi x}-1}\,dx=\frac16+\frac{\ln 2}{6}-2 \ln A \qquad \blacksquare$$
For the LogGamma integral between $0$ and $1/2$
Recall Kummer´s fourier expansion for LogGamma $0<x<1$
$$\ln\left(\Gamma(x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k x\right) + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k x \right) \tag{A.1}$$
Integrating we obtain
$$\int_0^x\ln\left(\Gamma(u)\right)du=\frac{x \ln\left(2 \pi\right)}{2}+\frac{1}{4 \pi}\sum_{k=1}^{\infty}\frac{\sin\left(2 \pi k x \right)}{k^2} +\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\cos \left(2 \pi k x \right)}{k^2}+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right)}{k^2}-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right) \cos\left(2 \pi k x \right)}{k^2} \tag{A.2}$$
Letting $x=\frac12$ in $(A.2)$ we obtain
$$\begin{aligned} \int_0^{1/2}\ln\left(\Gamma(x)\right)dx&=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{2 \pi^2}\zeta(2)+\frac{\gamma}{2 \pi^2}\eta(2)+\frac{\ln(2\pi)}{2 \pi^2}\zeta(2)+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k\right)}{k^2}+\frac{\ln \left( 2 \pi \right)}{2 \pi^2}\eta(2)-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k \right) \left(-1 \right)^{k}}{k^2}\\ &=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{12 }+\frac{\gamma}{24}+\frac{\ln(2\pi)}{12 }-\frac{1}{2 \pi^2}\zeta^{\prime}(2)+\frac{\ln \left( 2 \pi \right)}{24}-\frac{1}{2 \pi^2}\eta^{\prime}(2)\\ &=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{1}{2 \pi^2} \cdot \frac{\pi^2}{6}\left(\ln(2 \pi)+\gamma-12 \ln A \right)-\frac{1}{2 \pi^2} \left(\frac{\pi^2 \ln(2 )}{12} +\frac{\pi^2}{12}\left(\ln(2 \pi)+\gamma-12 \ln A\right)\right)\\ &=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{\ln(2 \pi)}{12}-\frac{\gamma}{12}+\ln A-\frac{\ln (2)}{24}-\frac{\ln(2 \pi)}{24}-\frac{\gamma}{24}+\frac{\ln A}{2}\\ &=\left(\frac{1}{8}-\frac{1}{12}-\frac{1}{24} \right)\gamma+\left(\frac38-\frac{1}{12}-\frac{1}{24} \right)\ln(2 \pi)-\frac{\ln(2)}{24}+\ln A+\frac{\ln A}{2}\\ &=\frac{3}{2}\ln A+ \frac{6}{24}\ln (2 \pi)-\frac{\ln (2)}{24}\\ &=\frac{3}{2}\ln A+\frac{5}{24}\ln 2+\frac{1}{4}\ln \pi \qquad \blacksquare \end{aligned}$$
We used that
$\zeta^{\prime}(2)=\frac{\pi^2}{6}\left(\ln 2 \pi + \gamma -12 \ln A \right)$
And that
$\eta^{\prime}(s)=2^{1-s}\ln (2) \zeta(s)+(1-2^{1-s})\zeta^{\prime}(s)$
Best Answer
You may have a look at the book Modern Analysis of Whittaker and Watson as cited in the Wikipedia article of the Digamma function. In Section 12.3 a proof is given by using the identity $$\tag{1}\frac{\Gamma'(z)}{\Gamma(z)} = -\gamma - \frac{1}{z} + \lim_{n \rightarrow \infty} \sum_{m=1}^n \Bigg( \frac{1}{m} - \frac{1}{z+m} \Bigg)$$ together with $$\frac{1}{z+m} = \int_0^\infty e^{-t(z+m)} \ dt.$$ In fact, this allows us to rewrite formula (1) by $$\frac{\Gamma'(z)}{\Gamma(z)} = -\gamma - \int_0^\infty e^{-zt} \ dt + \lim_{n \rightarrow \infty} \int_0^\infty \sum_{m=1}^n (e^{-mt} - e^{-(z+m)t}) \ dt.$$ Now the sum in the integral can be simplified by using the geometric series identity. In addition, we recall that $$\gamma = \int_0^\infty \Bigg(\frac{1}{1-e^{-t}} - \frac{1}{t} \Bigg) e^{-t} \ dt. $$ A proof of this identity is also given in the above-mentioned book. Combining both together, we see that \begin{align} \frac{\Gamma'(z)}{\Gamma(z)} &= -\gamma - \lim_{n \rightarrow \infty} \int_0^\infty \frac{e^{-t}-e^{-zt}-e^{-(n+1)t}-e^{-(z+n+1)t}}{1-e^{-t}} \ dt \\ & = \int_0^\infty \frac{e^{-t}}{t} - \frac{e^{-zt}}{1-e^{-t}} dt - \lim_{n \rightarrow \infty} \int_0^\infty \frac{1-e^{-zt}}{1-e^{-t}} e^{-(n+1)t} \ dt. \end{align} By continuous continuation we know that $\frac{1-e^{-zt}}{1-e^{-t}}$ is a bounded function on $[0,1]$. Moreover, for $t \ge 1$ one has obviously that $$ \bigg| \frac{1-e^{-zt}}{1-e^{-t}} \bigg| \leq \frac{2}{1-e^{-1}}$$ and therefore the function is bounded. Using this information, we can apply Lebesgue's theorem on dominated convergence to conclude that the integral vanishes for $n \rightarrow \infty$. All in all, we obtain that $$\frac{\Gamma'(z)}{\Gamma(z)} = \int_0^\infty \frac{e^{-t}}{t} - \frac{e^{-zt}}{1-e^{-t}} dt.$$ However, I am unaware which proof was given by Gauss and it is also not mentioned in the book.