Recent content by MountEvariste

Source: Yaghoub Sharifi.

Calculate $\displaystyle \int_0^{\infty} \frac{\sin x}{\cos x + \cosh x}\, \mathrm dx.$

Your answer and workings are correct. See here.

I don't know what you mean. It's true for $n=1$ because $10$ divides $120$.

$a \mid b$ 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...

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By the way, here's an interesting in-depth article in science magazine on Sweden's approach. I would love to know what you think @Ackbach

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...

Here.

Have you looked at Gilbert Strang lectures on YouTube?

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...