I managed to find$$\int\limits_0^\infty \frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x=-\pi\lim_{m\to \frac12 }\frac{d^{2a}}{d m^{2a}} \frac{\psi(1-m) + \gamma}{\sin(m\pi)}.$$

To show this relation, we follow the same approach here:

Reduce $n$ by $m$ in the beta function: $$\int_0^\infty\frac{x^{m-1}}{(1+x)^{m+n}}\mathrm{d}x=\operatorname{B}(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)},$$ we have $$\int_0^\infty\frac{x^{m-1}}{(1+x)^{n}}\mathrm{d}x=\frac{\Gamma(m)\Gamma(n-m)}{\Gamma(n)}.$$ Differentiate both sides $2a$ times with respect to $m$ and once with respect to $n$, \begin{gather*} \frac{\partial^{2a}}{\partial m^{2a}} \frac{\partial}{\partial n} \frac{\Gamma(m)\Gamma(n-m)}{\Gamma(n)}=\frac{\partial^{2a}}{\partial m^{2a}} \frac{\partial}{\partial n}\int\limits_0^\infty \frac{x^{m-1}}{(1+x)^n}\mathrm{d}x\\ \{\text{use differentiation under the integral sign theorem}\}\\ =\int\limits_0^\infty \frac{\partial^{2a}}{\partial m^{2a}} \frac{\partial}{\partial n}\frac{x^{m-1}}{(1+x)^n}\mathrm{d}x\\ =-\int\limits_0^\infty \frac{\ln^{2a}(x)\ln(1+x)x^{m-1}}{(1+x)^n}\mathrm{d}x. \end{gather*} Now take the limit on both sides letting $m\to 1/2$ and $n\to1$, \begin{gather*} -\int\limits_0^\infty \frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x=\lim_{\substack{m\to 1/2 \\ n \to 1}}\frac{\partial^{2a}}{\partial m^{2a}} \frac{\partial}{\partial n} \frac{\Gamma(m)\Gamma(n-m)}{\Gamma(n)}\\ =\lim_{\substack{m\to 1/2 \\ n \to 1}}\frac{\partial^{2a}}{\partial m^{2a}}\Gamma(m)\left( \frac{\partial}{\partial n} \frac{\Gamma(n-m)}{\Gamma(n)}\right)\\ =\lim_{\substack{m\to 1/2 \\ n \to 1}}\frac{\partial^{2a}}{\partial m^{2a}} \Gamma(m)\left(\frac{\Gamma(n-m)[{\psi}(n-m) -\psi(n)]}{\Gamma(n)}\right)\\ \{\text{evaluate the limit when $n\to 1$ and use $\psi(1)=-\gamma$}\}\\ =\lim_{m\to 1/2 }\frac{\partial^{2a}}{\partial m^{2a}} \Gamma(m)\Gamma(1-m)[\psi(1-m) + \gamma]\\ \left\{\text{use $\Gamma(m)\Gamma(1-m)=\frac{\pi}{\sin(m\pi )}$}\right\}\\ \left\{\text{and write $\frac{\partial}{\partial m}$ as $\frac{d}{dm}$, since we have one variable left}\right\}\\ =\pi\lim_{m\to \frac12 }\frac{d^{2a}}{d m^{2a}} \frac{\psi(1-m) + \gamma}{\sin(m\pi)}. \end{gather*}

The question here is can we write the limit in terms of finite summation?


3 Answers 3


This problem deals with the relation

$$\int\limits_0^\infty \frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x=-\pi\lim_{m\to \frac12 }\frac{\mathrm{d}^{2a}}{\mathrm{d} m^{2a}} \frac{\psi(1-m) + \gamma}{\sin(m\pi)}\tag{*}$$

The solution to the problem of the OP (write the r.h.s. as a finite sum) is found in the section "finite sum" below.

Optionally, we also try to verify the relation $(*)$. The main task, the transformation of the integral to a sum (plus explicit terms), is done in the first section.

Transformation of the integral

EDIT 17.05.21

I just discovered that Mathematica solves the generating integral (version 8.0 immediately, 10.1 via the antiderivative)

$$\int_0^{\infty } \frac{x^z \log (x+1)}{x+1} \, dx=\pi \csc (\pi z) (\psi ^{(0)}(-z)+\gamma )$$

The powers of $\log(x)$ unter the integral can be generated by differentiation.


We wish to evaluate the integral

$$i = \int_0^{\infty}\frac{ \log(x)^{2a}\log(1+x)}{\sqrt{x} (1+x)}\,dx\tag{1}$$

Where $2a$ is specified implicitly in the OP as a positive integer.

Splitting the integration region we can write

$$i=i_1 + i_2\tag{2}$$


$$i_1 = \int_0^{1}\frac{ \log(x)^{2a}\log(1+x)}{\sqrt{x} (1+x)}\,dx\tag{3a}$$

$$i_2 = \int_1^{\infty}\frac{ \log(x)^{2a}\log(1+x)}{\sqrt{x} (1+x)}\,dx\tag{3b}$$

Now letting $x\to \frac{1}{y}$ in $i_2$ we can write (skipping some steps)

$$i_2 = (-1)^{2a} i_1 + i_3\tag{4}$$


$$\begin{align}i_3 & = \int_0^1 \frac{(-\log (y))^{2 a+1}}{\sqrt{y} (y+1)} \, dy \\ & = 2^{-2 (a+1)} \left(\zeta \left(2 a+2,\frac{1}{4}\right)-\zeta \left(2 a+2,\frac{3}{4}\right)\right) \Gamma (2 a+2)\end{align}\tag{5}$$

Observing now that

$$\frac{\log(1+x)}{1+x} = -\sum_{n=1}^{\infty} H_n (-x)^n\tag{6}$$

We can write

$$i_1 =i_{1s}:=(-2)^{2 a+1}\Gamma (2 a+1) \sum_{n=1}^{\infty}(-1)^n \frac{H_n}{(2 n+1)^{2 a+1}}\tag{7a}$$

so that finally the integral $(1)$ can be written as

$$i=\left((-1)^{2 a}+1\right) i_{1s}+i_3$$

Hence the task of the OP reduces write the following sum $s$ as a sum of a finite number of terms

$$s = \sum_{n=1}^{\infty} (-1)^n\frac{ H_n}{ (1+2n)^{2a+1}}\tag{8}$$

Observation: this sum is evaluated in [1] identity $(4.91)$ .

Finite sum

Here we show that the r.h.s. of $(*)$ can be written an as finite sum.

Starting directly from the formula in the OP, letting $2a = k$ I get for the first few limits these expressions (format {k, limit})

$$ \begin{array}{c} \left\{0,-\pi \left(\psi ^{(0)}\left(\frac{1}{2}\right)+\gamma \right)\right\} \\ \left\{1,\frac{\pi ^3}{2}\right\} \\ \left\{2,-\pi \left(\pi ^2 \psi ^{(0)}\left(\frac{1}{2}\right)+\gamma \pi ^2+\psi ^{(2)}\left(\frac{1}{2}\right)\right)\right\} \\ \left\{3,\frac{5 \pi ^5}{2}\right\} \\ \left\{4,-\pi \left(5 \pi ^4 \psi ^{(0)}\left(\frac{1}{2}\right)+5 \gamma \pi ^4+6 \pi ^2 \psi ^{(2)}\left(\frac{1}{2}\right)+\psi ^{(4)}\left(\frac{1}{2}\right)\right)\right\} \\ \left\{5,\frac{61 \pi ^7}{2}\right\} \\ \end{array}\tag{9}$$

Now we could try to guess the rule for the polygamma functions and write the powers of $\pi$ in terms of $\zeta$-functions. But it is easier to proceed systematically.

We expand the $k$-th derivative of the two factor expression into a finite binomial sum

$$(\frac{d}{dm})^k (A B) = \sum_{j=0}^{k}\binom{k}{j} A^{(k-j)} B^{(j)}\tag{10}$$


$$A = \psi (1-m)+\gamma, B = \frac{1}{\sin{\pi m}}\tag{11}$$

Now using

$$\frac{\partial \psi ^{(p)}(1-m)}{\partial m}=-\psi ^{(p+1)}(1-m)\tag{12}$$

and the partial fraction decomposition

$$\frac{\pi}{\sin (\pi m)}=\sum _{n=-\infty }^{\infty } \frac{(-1)^n}{m-n}\tag{13}$$

we obtain for the j-th derivative the following expression

$$\begin{align}\frac{\partial ^j}{\partial m^j}\left(\frac{\pi }{\sin (\pi m)}\right)|_{m\to \frac{1}{2}}=-2^{-j-1}j! \left((-1)^k+1\right) \\ \times \left(-\zeta \left(j+1,\frac{1}{4}\right)+\zeta \left(j+1,-\frac{1}{4}\right)+4^{k+1}\right)\end{align}\tag{14}$$

With $(10)$, $(11)$, $(12)$, and $(14)$ we have all ingredients to expand the r.h.s. of $(*)$ into a finite sum.

I obtain for the finite sum requested in the OP (letting $2a \to k$) the following expression

$$r(k) := -\pi\lim_{m\to \frac12 }\frac{\mathrm{d}^{k}}{\mathrm{d} m^{k}} \frac{\psi(1-m) + \gamma}{\sin(m\pi)} \\=U(k) + \sum_{j=1}^{k} \binom{k}{j} U(k-j) V(j)\tag{15}$$


$$U(p) = -\pi \left(\gamma \delta _{0,p}+(-1)^p \psi ^{(p)}\left(\frac{1}{2}\right)\right) \tag{15a}$$

$$\begin{align}V(p) =-\frac{1}{\pi} \left(2^{-p-1} \left((-1)^p+1\right) p! \\ \times \left(-\zeta \left(p+1,\frac{1}{4}\right)+\zeta \left(p+1,-\frac{1}{4}\right)+4^{p+1}\right)\right) \end{align}\tag{15b}$$

Here $\delta _{0,p}$ is the Kronecker delta.

The first few $r(k)$ after simplification (FullSimplify) are in the format ${k,r(k)}$

for even $k$

$$ \begin{array}{c} \{0,\pi \log (4)\} \\ \left\{2,\pi ^3 \log (4)+14 \pi \zeta (3)\right\} \\ \left\{4,5 \pi ^5 \log (4)+84 \pi ^3 \zeta (3)+744 \pi \zeta (5)\right\} \\ \left\{6,61 \pi ^7 \log (4)+1050 \pi ^5 \zeta (3)+11160 \pi ^3 \zeta (5)+91440 \pi \zeta (7)\right\} \\ \left\{8,1385 \pi ^9 \log (4)+23912 \pi ^7 \zeta (3)+260400 \pi ^5 \zeta (5)+2560320 \pi ^3 \zeta (7)+20603520 \pi \zeta (9)\right\} \\ \end{array}\tag{16a} $$

for odd $k$

$$ \begin{array}{l} \left\{1,\frac{\pi ^3}{2}\right\} \\ \left\{3,\frac{5 \pi ^5}{2}\right\} \\ \left\{5,\frac{61 \pi ^7}{2}\right\} \\ \left\{7,\frac{1385 \pi ^9}{2}\right\} \\ \left\{9,\frac{50521 \pi ^{11}}{2}\right\} \\ \end{array}\tag{16b} $$

This finalizes the solution of the problem in the OP.


The numbers $c_{n}$ appearing here for both even and odd numbers

$$\{1, 1, 5, 61, 1 385, 50 521, 2 702 765, 199 360 981, ... \}$$

can be found in the impressivly long entry https://oeis.org/A000364:

A000364 Euler (or secant or "Zig") numbers.

They are related to the Euler polynomials: $c_{n}=2^n E_{n}(\frac{1}{2})$


[1] Ali Shadhar Olaikhan, "An introduction to harmonic series and logarithmic integrals", April 2021, ISBN 978-1-7367360-0-5

  • $\begingroup$ Thank you Wolf for the efforts... I have already related this integral to the sum you showed but i am asking if the limit can be written in terms of "finite" sum. Sorry I typed "definite sum" instead of "finite sum". $\endgroup$ May 15, 2021 at 0:24
  • $\begingroup$ Thank you, Ali, I got it now, see my answer with the explicit solution. $\endgroup$ May 15, 2021 at 11:34
  • $\begingroup$ Very nice Wolf. I like the idea in (11). I will investigate to see where the error is $\endgroup$ May 15, 2021 at 20:52
  • $\begingroup$ Thank you, Ali. The problem comes from the summand with $j=0$. I have now carefully collected the ingredients so that is should be possibe to compile it without errors. I keep on doing this. $\endgroup$ May 15, 2021 at 23:58
  • $\begingroup$ I think you meant $a$ or $2a$ for $k$ in (15) based on the binomial theorem? $\endgroup$ May 16, 2021 at 8:30

$$I=2{\pi^{2a+1}}\ln2|E_{2a}|+(2a)!{\pi^{2a+1}}\sum_{k=1}^{a} \frac{|E_{2a-2k}|}{(2a-2k)!}{\pi^{-2k}}(2^{2k+1}-1)\zeta(2k+1)$$

Where $$E_{2a}$$ are the Euler numbers $$E_{0}=1, E_{2}=-1,E _{4}=5,E _{6}=-61$$ Example: $$\int\limits_0^\infty \frac{\ln^{4}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x=10{\pi^5}\ln2+84{\pi^3}\zeta(3)+744{\pi}\zeta(5)$$

  • $\begingroup$ Interesting results thank you. +1 $\endgroup$ May 16, 2021 at 17:00
  • $\begingroup$ +1 Very nice and compact solution. How did you derive it? Just a typographical suggestion write \pi with lower case p. $\endgroup$ May 18, 2021 at 14:36
  • 1
    $\begingroup$ @Dr.Wolfgang Hintze $\int_0^{\frac{\pi}{2}}(Log\tan (x))^{2n}\log(\cos x)dx=\frac{\pi}{4}\lim_{m\to 0 }\frac{\mathrm{d}^{2n}}{\mathrm{d}\ m^{2n}} \frac{\psi(\frac{1-m}{2}) - \psi(1)}{\cos(\frac{m\pi}{2})}.$ Apply Mac-Laurin for ${\psi(\frac{1-m}{2})}$ $\endgroup$
    – user178256
    May 19, 2021 at 21:02
  • $\begingroup$ Can you give a reference for the Euler numbers? $\endgroup$ Jun 27, 2021 at 15:30
  • $\begingroup$ @Ali Shadhar see Table of Integrals, Series, and Products Fifth Edition By I.S.Gradshteyn and I.M.Ryzhik page XXXI $\endgroup$
    – user178256
    Jun 29, 2021 at 21:45

Thanks to @user178256 for the reference for the Euler number $( E_k)$ which is the key to solve the problem.

In the question body, we showed that

$$\int\limits_0^\infty \frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x=-\pi\lim_{m\to \frac12 }\frac{d^{2a}}{d m^{2a}} \frac{\psi(1-m) + \gamma}{\sin(m\pi)}$$


$$\frac{d^a}{dm^a}(f*g)=\sum_{k=0}^a \binom{a}{k} \frac{d^k}{dm^k}f*\frac{d^{a-k}}{dm^{a-k}} g$$

we have

$$I=-\pi\sum_{k=0}^{2a} \binom{2a}{k} \lim_{m\to \frac12}\frac{d^{k}}{dm^k} (\psi(1-m)+\gamma)*\lim_{m\to \frac12}\frac{d^{2a-k}}{dm^{2a-k}} \csc(m\pi)$$

since we have

$$\lim_{m\to \frac12}\frac{d^{0}}{dm^0} \csc(m\pi)=1$$

$$\lim_{m\to \frac12}\frac{d^{1}}{dm^1} \csc(m\pi)=0$$

$$\lim_{m\to \frac12}\frac{d^{2}}{dm^2} \csc(m\pi)=\pi^2$$

$$\lim_{m\to \frac12}\frac{d^{3}}{dm^3} \csc(m\pi)=0$$

$$\lim_{m\to \frac12}\frac{d^{4}}{dm^4} \csc(m\pi)=5\pi^4$$ or

$$\lim_{m\to \frac12}\frac{d^{r}}{dm^r} \csc(m\pi)=|E_{r}|\pi^r,$$

we consider only the even-$k$ terms and so

$$I=-\pi\sum_{k=0}^{a} \binom{2a}{2k} \lim_{m\to \frac12}\frac{d^{2k}}{dm^{2k}} (\psi(1-m)+\gamma)*\lim_{m\to \frac12}\frac{d^{2a-2k}}{dm^{2a-2k}} \csc(m\pi)$$

$$=-\pi^{2a+1}\sum_{k=0}^{a} \binom{2a}{2k} \lim_{m\to \frac12}\frac{d^{2k}}{dm^{2k}} (\psi(1-m)+\gamma)*|E_{2a-2k}|\pi^{-2k}$$

separate the first term using $\psi(1/2)+\gamma=-2\ln(2)$, we have

$$I=2\ln(2)|E_{2a}|\pi^{2a+1}-\pi^{2a+1}\sum_{k=1}^{a} \binom{2a}{2k} \lim_{m\to \frac12}\frac{d^{2k}}{dm^{2k}} (\psi(1-m)+\gamma)*|E_{2a-2k}|\pi^{-2k}$$


$$\lim_{m\to \frac12}\frac{d^{2k}}{dm^{2k}} (\psi(1-m)+\gamma)=\psi^{(2k)}\left(\frac12\right)=-(2k)!(2^{2k+1}-1)\zeta(2k+1)$$

we finally get

$$I=2\ln(2)|E_{2a}|\pi^{2a+1}+(2a)!{\pi^{2a+1}}\sum_{k=1}^{a} \frac{|E_{2a-2k}|}{(2a-2k)!}{\pi^{-2k}}(2^{2k+1}-1)\zeta(2k+1).$$

Bonus: By using $$\frac{1}{(2n+1)^{2a+1}}=\frac{1}{(2a)!}\int_0^1 x^{2n}\ln^{2a}(x)\mathrm{d}x,$$ we have \begin{gather*} \sum_{n=1}^\infty\frac{(-1)^{n}H_n}{(2n+1)^{2a+1}}=\frac{1}{(2a)!}\int_0^1\ln^{2a}(x)\left(\sum_{n=1}^\infty H_n(-x^2)^n\right)\mathrm{d}x\\ =\frac{1}{(2a)!}\int_0^1\ln^{2a}(x)\left(-\frac{\ln(1+x^2)}{1+x^2}\right)\mathrm{d}x\\ =-\frac{1}{(2a)!}\int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x \end{gather*}

where \begin{gather*} \int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x =\left(\int_0^\infty-\int_1^\infty\right)\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x\\ =\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x-\underbrace{\int_1^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x}_{x\to 1/x}\\ =\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x-\int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x\\ +2\int_0^1\frac{\ln^{2a}(x)\ln(x)}{1+x^2}\mathrm{d}x\\ \left\{\text{add $\int_0^1\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x$ to both sides then divide by 2}\right\}\\ =\frac12\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x^2)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\ln^{2a+1}(x)}{1+x^2}\mathrm{d}x\\ \overset{x^2\to x}{=}4^{-a-1}\int_0^\infty\frac{\ln^{2a}(x)\ln(1+x)}{\sqrt{x}(1+x)}\mathrm{d}x-(2a+1)!\beta(2a+2). \end{gather*}


$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{(2n+1)^{2a+1}}=(2a+1)\beta(2a+2)-\frac{\ln(2)|E_{2a}|}{(2a)!}\left(\frac{\pi}{2}\right)^{2a+1}$$ $$-\frac12\left(\frac{\pi}{2}\right)^{2a+1}\sum_{k=1}^{a} \frac{|E_{2a-2k}|}{(2a-2k)!}{\pi^{-2k}}(2^{2k+1}-1)\zeta(2k+1).$$

In case the reader is currious about the Mathematica command of the Euler number $(E_r)$, it is EulerE[r]


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