# Methods to derive a closed form for $I_n=\int_0^\infty \frac{\ln^n(x+1)-\ln^n(x)}{x+1}dx$

I've stumbled onto this general integral that has closed form values for the $$n\in \Bbb{Z^+}$$ $$I_n=\int_0^\infty \frac{\ln^n(x+1)-\ln^n(x)}{x+1}dx$$

Obviously $$I_0=0$$ but higher values of $$n$$ yield interesting results. The first few are: $$I_1=\frac{\pi^2}{6}$$ $$I_2=0$$ $$I_3=\frac{7\pi^4}{60}$$ $$I_4=0$$ $$...$$ So it appears that $$I_{2n}=0$$ and that $$I_{2n+1}=C_{2n+1}\zeta\left(2(n+1)\right)$$ where $$C_{n}$$ is some rational number that is equal to $$0$$ when $$n\in2\Bbb{Z^+}$$. Can anyone explain why $$I_{2n}=0$$ and how one can derive $$I_{2n+1}$$?

• Nice problem! +1 I find $I_n = \left(2-4^{-n}\right) (2 n+1)! \zeta ((2 n+1)+1)$ Oct 17, 2023 at 10:11

$$I_n=\int_0^\infty\frac {\ln^n(x+1)-ln^n(x)}{x+1}dx=\int_1^\infty\frac{\ln^n(x)}{x(x+1)}dx-\int_0^1\frac{\ln^n(x)}{1+x}dx$$, using $$x+1 \to x$$ in the $$ln^n(x+1)$$ term.

Let $$y=\frac{1}{x}$$ in the first integral and it becomes $$\int_0^1\frac{(-\ln(y))^n}{1+y}dy$$.

Net result $$I_n=0$$ for even $$n$$ and $$I_n=-2\int_0^1\frac{\ln^n(x)}{1+x}dx$$ for odd $$n$$. This last integral can be found in Gradshteyn and Ryzhik Table of Integrals...

• Can you list what the value for the integral is, because I cannot find it in any tables. Or can you link to where the value is stated? May 2, 2019 at 1:15
• Nevermind, Wolfram can evaluate it and the values match up. Thank you for your answer! May 2, 2019 at 3:59
• From the Table I referenced: $\int_0^1\frac{(lnx)^{2n-1}}{1+x}dx=\frac{1-2^{2n-1}}{2n}\pi^{2n}|B_{2n}|$ where $B_{2n}$ are Bernoulli numbers. If you can access that table, look for section 4.271. $May 2, 2019 at 19:38 • I see, I also realized that the integral you had$I_n\$ reduced to can be evaluated by a simple substitution to yield the answer in terms of the Gamma and Dirichlet Eta Function. May 2, 2019 at 19:40

I have found a closed form for the integral in question

$$I(n) = \int_0^\infty \frac{\ln^n(x+1)-\ln^n(x)}{x+1}\,dx\tag{1}$$

with a method which quite is common but has not been used in the partial solutions provided here by others.

$$g(a) = \int_0^{\infty } \frac{(x+1)^a-x^a}{x+1} \, dx\tag{2}$$

from which the $$I(n)$$ can be generated by differentiating $$n$$ times with respect to $$a$$ at $$a=0$$

$$g(a)$$ can be calculated explicitly: letting $$x\to \frac{t}{1-t}$$ gives

$$\int_0^1 (1-t)^{-a-1} \left(1-t^a\right) \, dt = \pi \csc (\pi a)-\frac{1}{a} \tag{3}$$

But here we have a nice series expansion

$$\pi \csc (\pi a)-\frac{1}{a}=-2 a \sum _{k=1}^{\infty } \frac{(-1)^k}{k^2-a^2}\tag{4}$$

Expanding the summand into a geometric series

$$-2a \sum _{k=1}^{\infty } \frac{ (-1)^k}{k^2-a^2}=-2a\sum _{k=1}^{\infty } \frac{(-1)^k}{k^2 \left(1-\left(\frac{a}{k}\right)^2\right)}\\ = -2 a\sum _{k=1}^{\infty } \sum _{n=0}^{\infty }\frac{(-1)^k }{k^2}\left(\frac{a}{k}\right)^{2 n}=-2 \sum _{n=0}^{\infty } a^{2 n+1} \sum _{k=1}^{\infty } \frac{(-1)^k}{k^{2 n+2}}$$

and doing the $$k$$-sum gives this power series in $$a$$

$$g(a) = \sum _{n=0}^{\infty } \left(2-4^{- n}\right) a^{2 n+1} \zeta (2 n+2))\tag{5}$$

from which we find this closed expression for our integral:

$$I(2n) = \left(\frac{\partial}{\partial a}\right)^{2n} g(a)|_{a\to 0}=0\tag{6a}$$ $$I(2n+1) =\left(\frac{\partial}{\partial a}\right)^{2n+1} g(a)|_{a\to 0}= \left(2-4^{-n}\right) (2 n+1)! \zeta (2 n+2)\tag{6b}$$