Welcome to my blog, a place for mathematics, physics, programming, and more!
A quick way to make work of some challenging integrals is the following: Glasser’s Master Theorem: Let $f(x)$ be a Riemann integrable function over $(-\infty,\,\infty)$ then \[\displaystyle{\i...
Problem For $x$ such that $f(x)\neq0$, define the $*$-derivative of $f$ at $x$ as [f^*(x)=\lim_{h\to0}\left(\frac{f(x+h)}{f(x)}\right)^{\frac{1}{h}}.] What is the general solution to the $*$-dif...
Problem Evaluate [I=\int_{-1/2}^{1/2}\sqrt{1+x^2+\sqrt{1+x^2+x^4}}\,\mathrm dx.] Solution Before doing any calculus, we first need to denest the radical. We can try to find two polynomial funct...
Some time ago I was messing around and brewed up a cute little Wallis-type product for the golden ratio, [\begin{align}\phi=\frac{2\cdot3}{1\cdot4}\cdot\frac{7\cdot8}{6\cdot9}\cdot\frac{12\cdot13}...
Problem Evaluate [I=\int_0^\frac{\pi}{2}\frac{\left{\tan(x)\right}}{\tan(x)}\,\mathrm dx] where $\left\{x\right\}$ denotes the fractional part of $x$. Solution First let $x=\text{arctan}(u)$ s...
Problem Evaluate [I=\int_0^1\frac{1-x}{1+x}\cdot\frac{1}{\ln(x)}\,\mathrm dx.] Solution Define the function [I(s)=\int_0^1\frac{1-x}{1+x}\cdot\frac{x^s}{\ln(x)}\,\mathrm dx] so that $I(0)=I$ ...
Let the sequence $\left\{x_n\right\}_{n=1}^\infty$ be the positive solutions to $\tan(x)=x$ ordered by increasing magnitude and define the function [\zeta_t(s)=\sum_{n=1}^\infty\frac{1}{x_n^s}.] ...
Problem Find the value of [f(x)=\sum_{n=0}^\infty\frac{2^n x^{2^n-1}-2^{n+1}x^{2^{n+1}-1}}{1-x^{2^n}+x^{2^n}}] when $x=1/2$. Solution The big hint in this problem is that the summand is the lo...
Problem Evaluate the sum [S=\sum_{n=1}^\infty\frac{1}{n^3}\sum_{k=1}^nH_k] where $H_k=\sum_{m=1}^k 1/m$ is the kth harmonic number. Solution Using the same method as in the solution of AMM1219...
Problem Evaluate [S=\sum_{n=1}^\infty\frac{1}{n^2}\sum_{k=1}^n\frac{1}{k^2}.] Solution The formula for summation by parts tells us that if we define [B_n=\sum_{k=0}^n b_k] so that for every $...
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