New answers tagged special-functions
0
votes
How to calculate singular moduli $\alpha_{3,n}$ of Ramanujan' s "$q_{3}$" theory?
This question has a lovely connection to the Monster group. (But Ramanujan didn't know it as that is 1970s maths.)
Let $q = e^{2\pi i\tau}$ and $\tau = \sqrt{-d}$ or $\tau = \frac{1+\sqrt{-d}}2$. ...
- 55.7k
0
votes
Interesting integral related to the Omega Constant/Lambert W Function
Beautiful Integral!
Because
\begin{align}
\frac{1}{{\left( e^z - z \right)^2 + \pi^2}} &= \frac{1}{\left( e^z - z - i\pi \right)\left( e^z - z + i\pi \right)} \\
&= \frac{1}{2\pi i}\left( \...
- 5,339
2
votes
I need help finding the value of this limit. spe$\lim_{n \to \infty}4^n \psi^{(2)}(2^n)$
Let $2^n=t$ and consider the problem of
$$f(t)=t^2 \,\psi ^{(2)}(t)$$
Use the asymptotics (have a look here)
$$\psi ^{(2)}(t)=-\frac 1 {t^2}-\frac 1 {t^3}-\frac 1 {2t^4}+O\left(\frac{1}{t^6}\right)$$
- 270k
1
vote
I need help finding the value of this limit. spe$\lim_{n \to \infty}4^n \psi^{(2)}(2^n)$
This is a method by Noureddine Sma
let $x=2^n$
$$\lim_{n\to \infty}4^n\psi^{(2)}(2^n) = \lim_{x\to \infty} x^2\psi^{(2)}(x) $$
we have :
$$\frac{(m-1)!}{x^m}+\frac{m!}{2x^{m+1}} \le (-1)^{m+1}\psi^{(m)...
- 3,090
1
vote
Repeated integral of $\tan x$
If we adopt the second form and using the series of polylogarithm function, we find that:
$$
\begin{split}
\int \tan x \,dx & =-\ln(\cos x)+C_{0} \\
&=\ln(2)+ix+\mathrm{Li}_{1}(-e^{2ix})+C_{...
1
vote
What is the sum of the series $( \frac{1}{2} - \frac{1}{3 \times 1!} + \frac{1}{4 \times 2!} - \frac{1}{5 \times 3!} + \dots $)?
Making the problem more general, consider
$$S(x) =\sum_{n=1}^\infty \frac{x^{n-1}}{(n+1)\, (n-1)!} $$
$$x^2\,S(x)=\sum_{n=1}^\infty \frac{x^{n+1}}{(n+1)\, (n-1)!} $$ Differentiate
$$\big(x^2\,S(x)\big)...
- 270k
3
votes
Repeated integral of $\tan x$
You can use Cauchy's formula for repeated integration to reduce the evaluation to a single definite integral:
$$f_n(x)=\int_{0}^{x}\int_0^{x_{1}}\cdots\int_{0}^{x_{n-1}}\tan x_n\, dx_n dx_{n-1}\cdots ...
- 2,269
0
votes
Accepted
Finding a simple (integro) differential equation to invert $\,_2\text F_1(a,b;c;z)$
$\def \F{\text F}$
This question has been unanswered for some time, so here is one for it. Start with the hypergeometric differential equation:
$$z(1-z)y’’+(c-(a+b+1) z)y’-a b y=0,y(0)=1,y’(0)=\frac{...
- 12.6k
0
votes
Accepted
Show that $\mathcal{L}^{-1}\left\{\frac{\Gamma(n)}{s^n}e^{-\frac{2a}{s}}{}_0F_1\left(n,\frac{a^2}{s^2}\right)\right\}(1)={}_0F_1\left(n,-a\right)^2$
By trial and error, I found the inverse Laplace transform. Given $g$, it is easy to prove this function corresponds to the Laplace inverse of $G$, by expressing it in terms of the square of a Bessel ...
- 53
0
votes
Is $\sqrt{\left(\operatorname{Si}(x)-\frac\pi2\right)^2+\operatorname{Ci}(x)^2}$ completely monotone?
Too long and too late for comments
What @SangchulLee wrote in comments is very interesting since
$$\Big[f(x)\Big]^2 = \int_{0}^{\infty} \mathrm{e}^{-xt}\, \frac{\log(1+t^2)}{t} \, \mathrm{d}t=\frac{1}{...
- 270k
1
vote
What is the closed form of $\sum_{n\geq 1}(-1)^{n-1}\psi'(n)^2$?
Solution: As we know, the integral representation of the trigamma function is
$$
\psi^{(1)}(k) = \int_0^1 \frac{x^{k-1} \ln(x)}{1-x} \, dx
$$
Using this, we get:
$$
\left( \psi^{(1)}(k) \right)^2 = \...
- 5,339
8
votes
Harmonic sum with Dirichlet eta tail
Here is my route in large steps
Consider the integral found in the OP
$$\eta \left(2\right)-\overline{H}_k^{\left(2\right)}=-4\int _0^1\left(-x^2\right)^k\frac{x\ln \left(x\right)}{1+x^2}\:dx,$$
as ...
- 1,516
0
votes
How to proceed in $I=\int_0^{\frac{\pi}{3}}\left(\ln\left(\frac{2\tan x}{\tan x +\sqrt{3}}\right)\right)^2\,dx$?
\begin{align}J&=\frac{\sqrt{3}}{2}\underbrace{\int_0^1\frac{\ln^2 x}{x^2-x+1}dx}_{=K}\\
K&\overset{u=\frac1x}=\int_1^\infty\frac{\ln^2u}{u^2-u+1}du=\frac{1}{2}\int_0^\infty\frac{\ln^2u}{u^2-u+...
- 14.3k
7
votes
Accepted
How to calculate the taylor series of $\int_{0}^{\arcsin{\frac{1}{x}}} \arccos{(x\sin{a})} da$?
Let us first take the substitution $a=\arcsin{t}$, so $da = \frac{1}{\sqrt{1-t^2}} dt$. This transforms your integral into:
$$
I(x) = \int_{0}^{\frac{1}{x}}\frac{\arccos{xt}}{\sqrt{1-t^2}} dt
$$
Now, ...
- 599
1
vote
Accepted
evaluating ${}_1F_1(a;b;x)$ and $\Gamma(x)$ at negative arguments
As noted in the answer above, $\Gamma(x)$ isn't well-behaved for non-positive integers. However, the regularized form of $_1F_1(a;b;z)$ does have a limiting form for negative integer values of $b$, ...
1
vote
2
votes
Accepted
Integrals of vector spherical harmonics
The orbital angular momentum operator in natural units ($\hbar=1$) is
$$\mathbf L=-i\,(\boldsymbol r\times\boldsymbol\nabla)=\mathrm L_x\boldsymbol i+\mathrm L_y\boldsymbol j+\mathrm L_z\boldsymbol k$$...
- 3,886
2
votes
Accepted
What is the origin of the non-analytic behavior of the integral $\int_0^\infty \frac{t^3}{t^4+1} J_2(x t) dt$?
Since no answers have been posted, I'm going to turn my comment into an answer.
I will show that $I(x)$ has the series expansion $$I(x) =\frac{1}{2} - \frac{\pi x^{2}}{32} + \frac{x^{4}}{384} \left(\...
- 43.1k
1
vote
How to proceed in $I=\int_0^{\frac{\pi}{3}}\left(\ln\left(\frac{2\tan x}{\tan x +\sqrt{3}}\right)\right)^2\,dx$?
Noticing that
$$
I=\frac{\sqrt{3}}{2} \int_0^1 \frac{\ln ^2 x}{x^2-x+1} d x= \frac{\sqrt{3}}{2} I^{\prime \prime}(0)
$$
where
$$
\begin{aligned}
I(a)&= \int_0^1 \frac{x^a}{x^2-x+1} d x \\
&= \...
- 23.8k
6
votes
Accepted
How to proceed in $I=\int_0^{\frac{\pi}{3}}\left(\ln\left(\frac{2\tan x}{\tan x +\sqrt{3}}\right)\right)^2\,dx$?
Note that
\begin{align}
K=&\int_0^1\frac{\ln^2 x}{x^2-x+1}dx\\
=&\int_0^1\frac{\ln^2 x}{x^3+1}dx
+ \int_0^1\frac{x\ln^2 x}{x^3+1}\overset{x\to 1/x}{dx}
=\int_0^\infty\frac{\ln^2x}{x^3+1}dx\\
=&...
- 102k
2
votes
How to proceed in $I=\int_0^{\frac{\pi}{3}}\left(\ln\left(\frac{2\tan x}{\tan x +\sqrt{3}}\right)\right)^2\,dx$?
In fact
$$\begin{eqnarray}
I&=&\frac{\sqrt3}{2}\int_0^1 \frac{\ln^2(x)}{x^2-x+1} dx=\frac{\sqrt3}{2}\int_0^1 \frac{(1+x)\ln^2(x)}{1+x^3} dx\\
&=&\frac{\sqrt3}{2}\int_0^1 \sum_{n=0}^\...
- 46k
15
votes
What is the sum of the series $( \frac{1}{2} - \frac{1}{3 \times 1!} + \frac{1}{4 \times 2!} - \frac{1}{5 \times 3!} + \dots $)?
$$\sum_{n=0}^\infty {(-1)^n\over n!(n+2)}=\sum_{n=0}^\infty {(-1)^n[(n+2)-1]\over (n+2)!}\\ =-\sum_{n=0}^\infty {(-1)^{n+1}\over (n+1)!}-\sum_{n=0}^\infty {(-1)^{n+2}\over (n+2)!}\\ =(1-e^{-1})-(-1+1-...
- 36.4k
3
votes
How to proceed in $I=\int_0^{\frac{\pi}{3}}\left(\ln\left(\frac{2\tan x}{\tan x +\sqrt{3}}\right)\right)^2\,dx$?
Rewrite $$I=\int\frac{\log ^2(x)}{x^2-x+1}\,dx=\int\frac{\log ^2(x)}{(x-a)(x-b)}\,dx$$where
$$a=\frac{1+i \sqrt{3}}{2}\qquad \text{and} \qquad b=\frac{1-i \sqrt{3}}{2} $$ Use partial fraction ...
- 270k
6
votes
What is the sum of the series $( \frac{1}{2} - \frac{1}{3 \times 1!} + \frac{1}{4 \times 2!} - \frac{1}{5 \times 3!} + \dots $)?
$$
\sum_{n\ge 0} \frac{(-)^n}{(n+2)n!}
=
\sum_{n\ge 0} \frac{(-)^n(2)_n}{2(3)_nn!}
=
\frac12 \sum_{n\ge 0} \frac{(-)^n(2)_n}{(3)_nn!}
=
\frac12 {}_1F_1(2;3;-1)
=1-2/e \approx 0.26424111
$$
a ...
- 3,606
24
votes
What is the sum of the series $( \frac{1}{2} - \frac{1}{3 \times 1!} + \frac{1}{4 \times 2!} - \frac{1}{5 \times 3!} + \dots $)?
Equals $$\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+2)}=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1 x^{n+1}\,dx=\int_0^1 xe^{-x}\,dx=1-\frac2e.$$
- 41k
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