4
votes
Accepted
Zeros of $\Gamma’$.
The gamma function has no zeros, therefore the zeros of its derivative are exactly the zeros of its logarithmic derivative, the digamma function, which has the series representation
$$
\psi (z)=-\...
- 119k
4
votes
Compute $ \xi_p= \prod_{n=1}^{\infty} (1+\frac {1}{n^p})$
lets start finding closed form for $\xi_n$ where
$$ \xi_p=\prod_{n=1}^\infty \left(1+\frac{1}{n^p} \right)=\prod_{n=1}^\infty \frac{n^p+1}{n^p} $$
and since
$$ n^p+1=0 \to n=e^{\frac{2k-1}{p}\pi i} , ...
- 1,827
4
votes
Good estimate for this the integral $\int_{\|x\|<R}e^{-\|x\|^2}dx_1\dots dx_m, x:=(x_1\dots x_m)$ in terms of $R,$ but possibly also $m?$
We know that
$$
\int_{\mathbb R^d} e^{-\lvert x \rvert^2}\, dx=\pi^\frac{d}{2}.$$
So the integral we want to estimate reads
$$
I_R=\pi^\frac{d}{2} - \int_{\lvert x\rvert> R} e^{-\lvert x \rvert^2}\,...
- 33k
4
votes
Good estimate for this the integral $\int_{\|x\|<R}e^{-\|x\|^2}dx_1\dots dx_m, x:=(x_1\dots x_m)$ in terms of $R,$ but possibly also $m?$
You can simply compute it. Via polar coordinates \begin{align*}
I(R)=\int_{B_R}e^{-\vert x \vert^2} \, dx &= \int_0^R \int_{\partial B_\rho} e^{-\vert x \vert^2} \, d\mathcal H^{n-1}_x \, d\rho\\
&...
- 7,574
3
votes
Accepted
Evaluate $\int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{dxdy}{\sqrt{\sin x\sin y(\cos x + \cos y)}}$
$$\begin{align*}
I &= \int_0^\tfrac\pi2 \int_0^\tfrac\pi2 \frac{dy\,dx}{\sqrt{\sin x \sin y (\cos x + \cos y)}} \\
&= \int_0^1 \int_0^1 \frac{\sqrt2\,dv\,du}{\sqrt u\sqrt v\sqrt{1-u^2v^2}} &...
- 21.2k
2
votes
Accepted
Gamma is multiplicative with Zeta - Is there an additive version?
Yes $$I(s)=2\sum_{n\in \Bbb N} \int_{J=(0,n^{-s/2})} e^{\frac{\log^2({n^{s/2}})}{\log t}}~dt$$
and $$ f(s)=\sum_{n\in \Bbb N} \log n~K_1(s\log n)$$
where $K_1$ is the modified Bessel function.
Here's ...
- 104
2
votes
Equation connecting The Riemann Zeta function and the Gamma function
If you write
$$\frac{x^{s-1}}{e^x+1}=\sum_{n=0}^\infty (-1)^n\,x^{s-1}\,e^{-(n+1) x}$$
$$I_n=\int_0^\infty (-1)^n\,x^{s-1}\,e^{-(n+1) x}\,dx=(-1)^n\, (n+1)^{-s} \,\Gamma (s)$$
$$\sum_{n=0}^\infty I_n=...
- 270k
1
vote
$\int_0^1 \sqrt{\frac{2-x^4}{1-x^4}}\, dx$ in terms of the gamma function or elliptic integrals
Too long for comment and not really an answer to the central question, but here is how we can recover the hypergeometric result:
$$\begin{align*}
I &= \int_0^1 \sqrt{\frac{2-x^4}{1-x^4}} \, dx \\
&...
- 21.2k
1
vote
Accepted
Integral derived from the correction factor of Sterling's approximation leading to a Puiseux series of polygammas
You are correct. Repeating your calculations
$$I=\int (1-\theta)\,dx=x-\frac 32 x^2-\frac {16} 3 x^3+4x^3\log(x)+3 x^2 \log (2 \pi x)+$$ $$12 \psi ^{(-3)}(x+1)-12 x \psi
^{(-2)}(x+1)$$
$$J=\int_0^...
- 270k
1
vote
Accepted
Asymptotic of incomplete Beta function
The median $\mathbf{m}$ of the beta distribution satisfies the equation
$$
\int_0^{\bf m} {u^{m - 1} (1 - u)^{n - 1} \mathrm{d}u} = \frac{1}{2}B(m,n).
$$
For $1 < m < n$, the median is bounded ...
- 33.6k
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