Mathematics Stack Exchange
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Questions tagged [digamma-function]

The digamma function, usually represented by the Greek letter psi or digamma, is the logarithmic derivative of the [tag:gamma-function]. It is the first of the polygamma functions.

$\psi(z)$ is defined as $\psi(z) =\frac{\mathrm{d}}{\mathrm{d}z}\log\Gamma(z)= \frac{\Gamma'(z)}{\Gamma(z)}.$

It has a series expansion converging everywhere except for the negative integers $\psi(z+1) = -\gamma + \sum_{n=1}^\infty \frac{z}{n(n+z)},$ where $\gamma$ is the Euler-Mascheroni constant.

$\psi(z)$ is a special case of the function $\psi^{(n)}(z)$ where $n=0$.

225 questions
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Beautiful monster: Catalan's constant and the Digamma function

The problem I have been trying for a while now to show that this monster $$\begin{align} &\int_0^{\pi/4}\tan(x)\sum_{n=1}^{\infty}(-1)^{n-1}\left(\psi\left(\frac{n}{2}\right)-\psi\left(\frac{n+1}{2}\right)+\frac{1}{n}\right)\sin(2nx)\,\mathrm{d}x…
vitamin d
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What is the closed form of $\sum_{n\geq 1}(-1)^{n-1}\psi'(n)^2$?

This problem was proposed by Cornel Ioan Valean. What is the closed form of $$ S=\sum_{n\geq 1}(-1)^{n+1}\psi'(n)^2 $$ ? I recall that $\psi'(z)=\frac{d^2}{dz^2}\log\Gamma(z)=\sum_{m\geq 0}\frac{1}{(m+z)^2}$ for any $z>0$. My attempt was to…
Jack D'Aurizio
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Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$

I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$ My Attempt: I start off by parametizing the integral as $$I(a)=\int_0^1…
aleden
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How to calculate $\int_0^{\frac{\pi}{2}}{x^2\cot x\ln\cos x\, \text{d}x}$.

Calculate: $$\int_0^{\frac{\pi}{2}}{x^2\cot x\ln\cos x\, \text{d}x}$$ My attempt: Let $$ A=\int_0^{\frac{\pi}{2}}{x^2\cot x\ln\cos x\, \text{d}x},B=\int_0^{\frac{\pi}{2}}{x^2\cot x\ln\sin x\, \text{d}x} $$ then $$ A+B=\frac{\pi ^2\ln…
Renascence_5.
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How to evaluate $\int_0^{\pi/2} x\ln^2(\sin x)\textrm{d}x$ in a different way?

The following problem \begin{align} &\int_{0}^{\pi/2} x\ln^{2}\left(\sin\left(x\right)\right)\,{\rm d}x \\[5mm] = & \ \frac{1}{2}\ln^{2}\left(2\right)\zeta\left(2\right) - \frac{19}{32}\,\zeta\left(4\right) + \frac{1}{24}\,\ln^{4}\left(2\right) +…
Ali Olaikhan
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Explicit series for the minimum point of the Gamma function?

Is there any explicit series, product, integral, continued fraction or other kind of expression for the point at which $\Gamma(x)$ has a minimum in $(0,1)$? The decimal value can be found here http://oeis.org/A030169, but I haven't found much…
Yuriy S
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Closed form to an interesting series: $\sum_{n=1}^\infty \frac{1}{1+n^3}$

Intutitively, I feel that there is a closed form to $$\sum_{n=1}^\infty \frac{1}{1+n^3}$$ I don't know why but this sum has really proved difficult. Attempted manipulating a Mellin Transform on the integral solution: $$\int_0^\infty…
user266519
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Conjecture for the value of $\int_0^1 \frac{1}{1+x^{p}}dx$

While browsing the post Is there any integral for the golden ratio $\phi$?, I came across this nice answer, $$ \int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\pi\,\phi}5$$ it seems the general form is just $$p \int_0^\infty…
Tito Piezas III
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Evaluation of $\sum_{n=1}^{\infty}\frac{1}{n(2n+1)}=2-2\ln(2)$

I came across the following statements $$\sum_{n=1}^{\infty} \frac{1}{n(2 n+1)}=2-2\ln 2 \qquad \tag{1}$$ $$\sum_{n=1}^{\infty} \frac{1}{n(3 n+1)}=3-\frac{3 \ln 3}{2}-\frac{\pi}{2 \sqrt{3}} \qquad \tag{2}$$ $$\sum_{n=1}^{\infty} \frac{1}{n(4…
Ricardo770
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Closed form of the sum $\sum_{n=1}^{\infty}\frac{H_n}{n^x}$

Some days ago I derived the identity $$\sum_{n=1}^{\infty}\frac{H_n}{n^2}=2\zeta(3)$$ where $H_n$ is the $n$th Harmonic number. Other related identities…
russian bot
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Is it possible to evaluate $\int_{-\infty}^{+\infty} x^n e^{-e^x+a x} dx $ for $a>0$?

The integral $$I=\int _{-\infty }^{+\infty }e^{-e^x+ax}dx \quad \textrm{ where }a>0 $$ seems very difficult. However, putting $y=e^x$ converts the integral into $$ I=\int_0^{+\infty} y^{a-1} e^{-y} d y=\Gamma(a) $$ Then I try to generalise it…
Lai
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How to prove $\sum_{k=1}^n{n\choose k}\frac{(-1)^{k+1}}{k}=\sum_{k=1}^n\frac{1}{k}$?

I think by induction we can do it. Let $I(n)=\sum_{k=1}^n{n\choose k}\frac{(-1)^{k+1}}{k}=\sum_{k=1}^n\frac{1}{k}.$ Then, we must show that $I(n+1)-I(n)=\frac{1}{n+1}$. $\begin{align}…
Bob Dobbs
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Compute the double sum $\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)$

I am trying to compute the following double sum $$\boxed{\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)}$$ I proceeded as following $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2…
Ricardo770
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What do we know about $\sum_\limits{n=0}^{\infty} \frac{(-1)^n}{kn+1}$?

Let define, for $k \ge 1$ : $$ f(k) = \sum_\limits{n=0}^{\infty} \frac{(-1)^n}{kn+1}. $$ It is well-known that $f(1) = \ln(2), f(2) = \pi/4$. Some computations on WolframAlpha led me to $f(3) =1/9 (\sqrt3 \pi+\ln(8))$, $f(4) = (\pi+2…
Watson
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Integral of principal value of $[\tanh(x+a)-\tanh(x+b)]/x$

How do you solve this integral involving the Cauchy principal value? $$ \mathcal{P} \int_{-\infty}^{\infty} \frac{\tanh(x+a)-\tanh(x+b)}{x} dx \\ = \int_0^\infty \frac{\tanh(x+a)+\tanh(x-a)-\tanh(x+b)-\tanh(x-b)}{x} dx $$ The integral converges as…
Bio
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