All Questions
Tagged with harmonic-numbers integration
259 questions
0
votes
4
answers
231
views
Integrating $\int_0^1 \frac{\ln x \ln (1+x) \ln(1-x)}{x^{\frac12}} \,dx$
$$\int_0^1 \frac{\ln x \ln (1+x) \ln(1-x)}{x^{\frac12}}
\,dx$$
Now I have seen a similar integral here,
I am unable to deal with the $x^{\frac12}$ term in the denominator.
Here's one idea,
$$\int_0^1x^...
- 5,658
6
votes
0
answers
177
views
How to evaluate $-\sum_{n=0}^{\infty} \left( \frac{4^n}{\left( (2n+1) \binom{2n}{n} \right)^2} \right) \frac{H_{n+1}}{n+1}$
Question
How to evaluate
$$
-\sum_{n = 0}^{\infty}
\frac{4^{n}}{\left[\rule{0pt}{5mm}
\left(2n + 1\right)\binom{2n}{n}\right]^{2}}\
\frac{H_{n + 1}}{n +1}
$$
My attempt
Due to another shifted sum
$$
...
- 5,967
1
vote
1
answer
110
views
how to evaluate $\sum_{n=1}^\infty \sum_{m=1}^{n-1} \frac{(-t)^{m+n}}{m-n}$?
I tried to find the summation
$$ \Omega=\sum_{n=1}^\infty \sum_{m=1 , m\ne n}^\infty \frac{(-1)^{m+n} H_{m+n}}{m^2-n^2}$$
and got
$$ \Omega=\int_0^1 \frac{\ln(1-t)}{t} \left( \frac{t^2}{1-t^2} \ln(1+t)...
- 1,979
1
vote
1
answer
164
views
How to evaluate $\sum_{n=1}^{\infty}\frac{H_{n}^{(2)}}{n^{4}}$
It may be rather tedious and I will have to delve into deeper, but I have a little something. We probably already know this one. The thing is, the first one results in yet another Euler sum. But, I ...
- 5,967
5
votes
2
answers
166
views
Show that $\sum_{n=1}^{\infty} \frac{\binom{2n}{n} (H_{2n} - H_n)}{4^n (2n - 1)^2} = 2 + \frac{3\pi}{2} \log(2) - 2G - \pi$
Show that $$\sum_{n=1}^{\infty} \frac{\binom{2n}{n} (H_{2n} - H_n)}{4^n (2n - 1)^2} = 2 + \frac{3\pi}{2} \log(2) - 2G - \pi$$
My try :
We know that
$$\sum_{n=1}^{\infty} \binom{2n}{n} (H_{2n} - H_{n}) ...
- 5,967
7
votes
1
answer
462
views
Ramanujan-Type Harmonic Series $\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{1/6+r}{2^{8r}}\right)H_{r-1/2}=-\frac{8\ln2}{9\pi}$
We will be considering series such as this :
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{1/6+r}{2^{8r}}\right)H_{r-1/2}=-\frac{8\ln2}{9\pi}$$
Consider $K(k)$ and $E(k)$ to be the Complete ...
- 3,710
4
votes
2
answers
456
views
Harmonic sum with Dirichlet eta tail
The following problem is proposed by Cornel Valean:
$$\sum_{n=1}^{\infty} \frac{H_n}{2n+1}\left(\eta(2)- \overline{H}_n^{(2)}\right)$$
$$=2 G^2-2\ln(2) \pi G+\ln^2(2)\frac{\pi ^2}{6} +\frac{53}{1440}\...
- 26.6k
8
votes
1
answer
2k
views
Proving $\sum_{n=1}^{\infty}\binom{2n}{n}^2\frac{4H_{2n}-3H_n}{n2^{4n}}=\zeta(2)$
While trying to solve Prove $\int_{0}^{1}\frac1k K(k)\ln\left[\frac{\left(1+k \right)^3}{1-k} \right]\text{d}k=\frac{\pi^3}{4}$
I was able to reduce it to the following form,
$$\sum_{n=1}^{\infty}\...
- 3,710
11
votes
0
answers
286
views
Solve the integral $\int_0^1 \frac{\ln^2(x+1)-\ln\left(\frac{2x}{x^2+1}\right)\ln x+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx$
I tried to solve this integral and got it, I showed firstly
$$\int_0^1 \frac{\ln^2(x+1)+\ln^2\left(\frac{x}{x+1}\right)}{x^2+1} dx=2\Im\left[\text{Li}_3(1+i) \right] $$
and for other integral
$$\int_0^...
- 1,979
2
votes
1
answer
192
views
On the evaluation of $\int_{0}^{1}x^{n-1}\log(1+x)\,dx$ in terms of the harmonic numbers.
Is my evaluation of this integral correct? Here is my work.
$$I:=\int_{0}^{1}x^{n-1}\log(1+x)\,dx$$
$I.B.P$
$$I=\frac{1}{n}\log(2)-\frac{1}{n}\int_{0}^{1}\frac{x^n}{1+x}\,dx$$
Define:
$$J:=\int_{0}^{1}...
- 1,163
5
votes
2
answers
257
views
Does anyone have good logarithmic integrals? And logarithmic integral identities?
I have recently taken an interest in evaluating logarithmic integrals and would really love practice problems and especially, theorems, series expansions, and identities that have helped any of ya’ll ...
- 1,163
17
votes
2
answers
567
views
Help with $\sum _{n=1}^{\infty }\frac{H_{2n}^2\binom{2n}{n}}{4^nn^2}$
I want to know what strategies I can use in order to evaluate:
$$\sum _{n=1}^{\infty }\frac{H_{2n}^2\binom{2n}{n}}{4^nn^2}.$$
What I tried is the following:
$$2\int _0^1x^{2n-1}\ln ^2\left(1-x\right)\:...
- 240
8
votes
0
answers
411
views
Two nice (challenging) binoharmonic series
I've recently seen a nice binoharmonic series from Ali's book (page 309), but that post, for some reason, vanished (Update: that post was undeleted and now it may be found here Evaluating $\int_0^{\...
- 5,632
17
votes
3
answers
671
views
Evaluating $\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x$
I was able to find
$$\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x=-\frac14\sum_{n=1}^\infty\frac{4^n}{{2n\choose n}}\frac{H_{2n}}{n^3}$$
$$=5\operatorname{Li}_4\left(\frac12\right)-\frac{...
- 26.6k
1
vote
1
answer
87
views
$\log{n} + \frac{1}{n} < H_n < \log{n} + 1$
I see on that this post that the inequality $\log{n} < H_n < \log{n} + 1$ is proven. However, Proofs from THE BOOK takes it a step further, and states that $\log{n} + \frac{1}{n} < H_n < \...
- 356
4
votes
1
answer
116
views
Prove that the limit converges to $\gamma$
Let $$H(x)=\int_0^1\frac{t^x-1}{t-1}dt$$be the harmonic series and let $$s(x)=\int^\infty_0e^{-t}\ln(t+x)dt$$How do I prove that their difference converges? It seems to me that they approach $\gamma$ (...
- 6,697
-2
votes
1
answer
64
views
An interesting integral about γ [closed]
It is well known that
$$\gamma = \lim_{n \to +\infty} (H_n − \ln(n)) = 0.5772(...)$$
where $H_n$ is the sum of reciprocals of all integers from $1$ to $n$.
Prove that $$\int_1^\infty \frac{\{ x\}}{x^...
7
votes
4
answers
474
views
Integral Involving Harmonic Numbers: $\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx$
(Motivation) In an attempt to answer this question, I got stuck on evaluating a certain integral. I have made up the following conjecture:
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}...
- 5,813
3
votes
1
answer
159
views
Find a closed form for the sum with harmonic number $\sum_{n=0}^\infty (-1)^{n}\frac{H_{2n+1}^{2}}{(2n+1)^2}$
In this link gave an answer for a question which invoved the sum of generalized harmonic numbers with power $2n+1$.
And I thought about this sum:
$$S=\sum_{n=1}^\infty (-1)^{n}\frac{H_{2n+1}^{2}}{(2n+...
- 2,887
11
votes
0
answers
441
views
Is the closed form of $\int_0^1\frac{\text{Li}_{2a+1}(x)}{1+x^2}dx$ known in the literature?
Using
$$\text{Li}_{2a+1}(x)-\text{Li}_{2a+1}(1/x)=\frac{i\,\pi\ln^{2a}(x)}{(2a)!}+2\sum_{k=0}^a \frac{\zeta(2a-2k)}{(2k+1)!}\ln^{2k+1}(x)\tag{1}$$
and
$$\int_0^1x^{n-1}\operatorname{Li}_a(x)\mathrm{d}...
- 26.6k
3
votes
2
answers
145
views
Show that $\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$
How to show that
$$\int_0^1\frac{\ln(1-x)}{1+x^2}\mathrm{d}x+\int_0^1\frac{\arctan x}{x(1+x)}\mathrm{d}x=0$$
without evaluating each integral individually?
I came up with this problem while working on ...
- 26.6k
6
votes
4
answers
515
views
Finding $\sum_{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$
I want to find the closed form of:
$\displaystyle \tag*{} \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$
Where $H_{k}$ is $k^{\text{th}}$ harmonic number
I tried to ...
- 951
5
votes
1
answer
280
views
Are these two recursive formulas known in the literature?
First let me introduce the two recursive relations:
$$\int_0^1 x^{n-1}\ln^a(1-x)dx=f(a,n),$$
where
$$f(a,n)=(a-1)!\sum_{j=0}^{a-1}\frac{(-1)^{a-j}}{j!}f(j,n) H_n^{(a-j)},\quad f(0,n)=\frac1n.\tag{1}$$...
- 26.6k
37
votes
0
answers
2k
views
Are these generalizations known in the literature?
By using
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\tag{a}$$
and
$$\text{Li}_{a}(-z)+(-1)^a\text{Li}_{a}(-1/z)=-2\sum_{k=0}^{\lfloor{a/2}\rfloor }\frac{\eta(...
- 26.6k
3
votes
1
answer
104
views
Change in order of integration in an integral
I do not understand the change of order of integration ($dtdx$ to $dxdt$, RHS second line). Why the $\frac{1}{2}$ term? Why do we integrate $x$ from $t$ to $1$?
- 55
7
votes
0
answers
114
views
Convergence of double integral involving logarithms and binomials? Can anyone present a simpler proof?
I am working through a paper I posted to arXiv and am looking to condense my results as well as seek alternative proofs for some of my intermediate results. In particular, I am looking for a shorter ...
- 6,279
9
votes
1
answer
334
views
Different ways to evaluate $\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$
The following question:
How to compute the harmonic series $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$$
where $H_n=\sum_{k=1}^n\frac{1}{k}$ and $H_n^{(2)}=\sum_{k=1}^n\frac{1}{k^2}$, was ...
- 26.6k
0
votes
1
answer
171
views
Is there a closed a form for $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{an}}{n}\,?$
I am wondering if there is a closed form for
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{an}}{n},$$
Where $a$ is positive real.
The integral representation of this sum is
$$a\zeta(2)+a\int_0^1\frac{\ln(1-x)}...
- 26.6k
3
votes
0
answers
322
views
Fascinating equality
The following problem is proposed by a friend:$$4\int_0^{\pi/4}\left(\int_x^{\pi/4} (x-y)\ln(\tan (x)) \ln\left(\tan \left(y+\frac{\pi }{4}\right)\right) \textrm{d}y\right)\textrm{d}x$$
$$-\sum _{n=1}^...
- 26.6k
10
votes
2
answers
676
views
Evaluating $\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x$
While I was working on computing $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}$, I came across the integral:
$$I=\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x.$$
I tried ...
- 26.6k
7
votes
3
answers
291
views
Compute the double sum $\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)$
I am trying to compute the following double sum
$$\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}^...
- 2,871
3
votes
4
answers
485
views
How to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}H_{2n}$
Edit
In this post I computed the following integral
$$\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx=\frac{11}{8}\zeta(3)$$
Now I am trying to compute
$$\boxed{\int_{0}^{1}\frac{\log(1-x)\log(1-x^4)}{x}...
- 2,871
5
votes
1
answer
270
views
Where is my mistake $\int_{0}^{1}\frac{\log(1-x)\log(1-x^2)}{x}dx$
Edit
As commented bellow by @Donald Splutterwit and @ Elliot Yu, it seems that my computation is numerically correct and the post is wrong! I also added a corollary from this computation.
I saw the ...
- 2,871
3
votes
0
answers
158
views
Evaluating the integral: $\int _0^1\frac{\ln \left(x\right)\ln ^2\left(\frac{1-x}{1+x}\right)\operatorname{Li}_2\left(x\right)}{x}\:dx$
I'd like to evaluate and find the closed form for
$$\int _0^1\frac{\ln \left(x\right)\ln ^2\left(\frac{1-x}{1+x}\right)\operatorname{Li}_2\left(x\right)}{x}\:dx$$
But it's very difficult since its ...
user809806
6
votes
4
answers
263
views
Closed form of $\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$
I've been working on an integral, namely: $${\displaystyle \int_0^1 \frac{x^2}{1 + x^3}\ln(1 - x^4)dx}$$
Which I managed to narrow down to the following expression: $$\sum_{m=1}^{\infty} \frac{(-1)^...
1
vote
1
answer
99
views
Proving an equality involving harmonic numbers
I am working through the book (Almost) Impossible Integrals, Sums, and Series, and I am on the following problem:
Let $n$ be a positive integer. Prove that the following equality holds.
$$K_n = \...
- 165
1
vote
1
answer
165
views
How to Evaluate $ \sum_{n=1}^{\infty} \frac{(-1)^n}{n}H_{4n-3} $
I came across the following result:
$$ \sum_{n=1}^{\infty} \frac{(-1)^n}{n}H_{4n-3} = \frac{\pi}{4} + \frac{\pi}{\sqrt{2}}-\frac{5\pi^2}{32}+\frac{\ln^2(2)}{8}-\frac{3\ln(2)}{2}+\frac{\ln^2(1+\sqrt{2})...
- 1,236
4
votes
1
answer
505
views
How to evaluate $\sum _{n=1}^{\infty }\left(\frac{H_n^2+H_n^{\left(2\right)}}{n}\right)^2$ in a particular way.
How to evaluate:
$$\sum _{n=1}^{\infty }\left(\frac{H_n^2+H_n^{\left(2\right)}}{n}\right)^2,$$
without splitting the expression into more sums.
Here $H_n^{\left(m\right)}=\sum _{k=1}^n\frac{1}{k^m}$ ...
user809806
15
votes
2
answers
812
views
Evaluating the challenging sum $\sum _{k=1}^{\infty }\frac{H_{2k}}{k^3\:4^k}\binom{2k}{k}$.
I managed to evaluate the sum, my approach can be found $\underline{\operatorname{below as an answer}}$, I'd truly appreciate if any of you could share new methods to evaluate this series, thank you.
...
- 2,683
10
votes
2
answers
619
views
Evaluating $\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$.
My attempt.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$$
$$=\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k}\binom{2k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\...
- 2,683
9
votes
4
answers
348
views
Alternative ways to evaluate $\int_0^1 x^{2n-1}\ln(1+x)dx =\frac{H_{2n}-H_n}{2n}$
For all, $n\geq 1$, prove that
$$\int_0^1 x^{2n-1}\ln(1+x)\,dx =\frac{H_{2n}-H_n}{2n}\tag1$$
This identity I came across to know from here, YouTube which is proved in elementary way.
Trying to make ...
- 3,482
6
votes
0
answers
326
views
Does there exist a closed form for $\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$?
I am not sure if there exists a closed form for
$$I=\int_0^{\pi/2}\frac{x^2\ \text{Li}_2(\sin^2x)}{\sin x}dx$$
which seems non-trivial.
I used the reflection and landen's identity, didn't help much.
...
- 26.6k
4
votes
2
answers
349
views
How to approach $\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}$ elegantly?
How to show that
$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac{\psi^{(3)}\left(\frac14\right)}{384}-\frac{\pi^4}{48}-\frac{35\pi}{128}\zeta(3)$$
without using the generating function:
\...
- 26.6k
2
votes
0
answers
217
views
How to evaluate $\int _0^1\frac{\ln ^2\left(x\right)\ln ^3\left(\frac{1-x}{1+x}\right)}{1-x}\:dx$.
How can I evaluate $\displaystyle \int _0^1\frac{\ln ^2\left(x\right)\ln ^3\left(\frac{1-x}{1+x}\right)}{1-x}\:dx$ in an elegant (simple?) way? this integral seems very challenging, by simply ...
user809806
6
votes
2
answers
746
views
Challenging integral: $\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\sin x}dx$
How to tackle
$$I=\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\sin x}dx\ ?$$
This integral popped up in my solution ( see the integral $\mathcal{I_3}\ $ at the end of the solution.)
My attempt: By ...
- 26.6k
7
votes
2
answers
1k
views
How to find $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$ using real methods?
How to calculate
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}$$
and
$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}^{(2)}}{n^2}$$
by means of real methods?
This question was suggested by Cornel the author of the ...
- 26.6k
6
votes
2
answers
386
views
Computing $\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{2n+1}$ in an alternative way
The following equality
$$\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^2}{2n+1}=\frac{3}{16}\pi^3-\frac34\ln^2(2)\pi-8\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
can be proved if we are allowed ...
- 26.6k
3
votes
1
answer
345
views
Resisting integral: $\int_0^1\frac{\arcsin^2(x)\ln(1-x)}{x}dx$
Following the same technique in this question , one can find
$$\sum_{n=1}^\infty\frac{4^nH_{2n}}{n^3{2n\choose n}}=-4\int_0^1\frac{\arcsin^2(x)\ln(1-x)}{x}dx=-4\int_0^{\pi/2}x^2\cot x\ln(1-\sin x)dx$$
...
- 26.6k
17
votes
2
answers
1k
views
How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$?
@User mentioned in the comments that
$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)\tag1$$
$$\small{\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi ...
- 26.6k
8
votes
4
answers
718
views
How to evaluate $\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$
Before you think I haven't tried anything, please read.
I've been trying to evaluate $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$$
But I can't find a way to simplify it. ...
user809806