User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

DiffEQ Ongoing Project
Assignment:


 * $\displaystyle \int_{\to 0}^{\to 1} \ln x \ln \left({1-x}\right)\ \mathrm dx$

To prove the integral exists, note that $\ln$ is continuous for all $x$ in its domain.

As $x \to 0^+$:

As $x \to 1^{-}$:

Find the power series of $\ln(1-x)$:

Then,

Let $n \in \N_{\ge 1}$

So,

Since $n$ was arbitrary, this holds for all cases $n \ge 1$, for $x \in (0\,.\,.\,1)$.

--GFauxPas (talk) 13:41, 25 March 2014 (UTC)

.... So I got to this:


 * $\displaystyle \int_{\to 0}^{\to 1} \ln x \ln \left({1-x}\right)\ \mathrm dx = \sum_{n \ge 1} \frac 1 {n(n+1)}$

How do I get past that? It's kind of like the Riemann Zeta function at 2...

--GFauxPas (talk) 14:48, 3 April 2014 (UTC)


 * $\frac1{n(n+1)} = \frac1n-\frac1{n+1}$. Hence $\sum_{n\ge1}^k \frac1{n(n+1)} = 1 - \frac1{k+1}$. Conclude. &mdash; Lord_Farin (talk) 16:51, 3 April 2014 (UTC)


 * Major typo, should have been $\sum_{n \ge 1} \frac 1 {n(n+1)^2}$. Nevertheless your hint did help me!

The integral ends up being $2-\zeta(2)$. --GFauxPas (talk) 18:52, 3 April 2014 (UTC)

Definition
Let:


 * $a \dfrac{\mathrm d^2y}{\mathrm dx^2} + b \dfrac {\mathrm dy}{\mathrm dx} + c y = f\left({ x,y,\dfrac {\mathrm d^2y}{\mathrm dx^2},\dfrac {\mathrm dy}{\mathrm dx}}\right)$

be a second order linear ODE.

The characteristic equation of this ODE is:


 * $ar^2 + br + c = 0$

where $y = e^{rx}$.

--GFauxPas (talk) 14:49, 3 April 2014 (UTC)