$a \mid b$ means $a$ divides $b$.
The following formula may help.
Legendre's formula: For any prime number $p$ and any positive integer $m$, let
${\displaystyle \nu _{p}(m)}$ be the exponent of the largest power of $p$ that divides $m$. Then
$${\displaystyle \nu _{p}(m!)=\sum _{i=1}^{\infty...
Be that as it may, could you explain why Sweden’s numbers would so vastly different to those of their closest neighbours (who happen to have had lockdown unlike Sweden)?
Well if you impose lockdown when the disease is already rampant in the community, lockdown isn’t going to be as effective as...
Letting $y \mapsto \frac{1}{y}$ we get $\displaystyle I = \int_0^1 \frac{y-1}{(y^3+1)\log{y}}\,\mathrm{d}y$; since $\displaystyle \frac{y-1}{\log{y}}= \int_0^1 y^t \, \mathrm{d}t$, we have
$$\begin{aligned} I & = \int_0^1 \int_0^1 \frac{y^t}{y^3+1}\,\mathrm{d}t\,\mathrm{d}y = \int_0^1 \int_0^1...
1. Sub $y \mapsto \frac{1}{y}$ to change the bounds.
2. Note that $\displaystyle \frac{y-1}{\log{y}}= \int_0^1 y^t \, \mathrm{d}t $.
3. Switch the order of the double integral
4. Expand the geometric series
5. Switch the order of integral and series
6. You're left with an infinite series...