All Questions
8
questions
17
votes
2
answers
894
views
A reason for $ 64\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt =-\pi^4$ ...
Question: How to show the relation
$$
J:=\int_0^1 \left(\frac \pi 4+\arctan t\right)^2\cdot \log t\cdot\frac 1{1-t^2}\; dt
=-\frac 1{64}\pi^4
$$
(using a "minimal industry" of relations, ...
- 34.2k
2
votes
2
answers
226
views
Evaluation of a log-trig integral in terms of the Clausen function (or other functions related to the dilogarithm)
Define the function $\mathcal{I}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{I}{\left(a,\theta\right)}:=\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(1-2a\cos{\left(\...
- 30.7k
1
vote
0
answers
145
views
Dilogarithm of a negative real number outside unit circle
The dilogarithm is defined in $\mathbb{C}$ as
$$
Li_2(z) = -\int_0^1 \frac{\ln(1 - zt)}{t} dt
$$
Because $1-zt \in \mathbb{C}$, then you can write $\ln(1 - zt) = \ln|1 - zt| + i·\arg(1 - zt)$
As ...
- 539
4
votes
2
answers
260
views
Logarithmic integral $ \int_0^1 \frac{x}{x^2+1} \, \log(x)\log(x+1) \, {\rm d}x $
At various places e.g.
Calculate $\int_0^1\frac{\log^2(1+x)\log(x)\log(1-x)}{1-x}dx$
and
How to prove $\int_0^1x\ln^2(1+x)\ln(\frac{x^2}{1+x})\frac{dx}{1+x^2}$
logarithmic integrals are connected ...
- 6,277
3
votes
2
answers
304
views
Jump of dilogarithm
I am reading about the dilogarithm function
$$ \mathrm{Li}_2(z):= - \int_0^z \frac{\log(1-u)}{u}du, \quad z \in \mathbb{C} \backslash [1, \infty).$$
I found it stated that the "jump" of the ...
- 6,326
3
votes
3
answers
449
views
Indefinite integral $\int \arctan^2 x dx$ in terms of the dilogarithm function
I read about the integral
$$\int \arctan^2 x dx$$ in this old post: Evaluation of $\int (\arctan x)^2 dx$
By replacing
$$\arctan x = -\frac{i}{2}\left[\log(1+ix) - \log(i-ix)\right],$$
as suggested ...
- 7,613
3
votes
1
answer
709
views
What's about $\int_0^\infty x^{-z}Li_{z+2}(e^{-xz})dx$ as $-\zeta(3)z^2\Gamma(-z)$?
Because $$-\int_0^\infty \frac{e^{-zt}}{t^z}dt=z^2\Gamma(-z),$$
holds for $0<\Re z<1$ then using the change of variable $t=nx$ one has that $$-\frac{1}{n^{z-1}}\int_0^\infty \frac{e^{-znx}dx}{x^...
user243301
0
votes
1
answer
406
views
Evaluating Fermi Dirac integrals of order j<0
The complete Fermi Dirac integral (I'm purposely leaving off the gamma prefactor)...
$$F_j(x) =\int\limits_{0}^{\infty} \frac{t^j}{e^{t-x}+1} \: dt$$
is generally defined for j>-1. Is there a way to ...
- 1