Welcome to my blog, a place for mathematics, physics, programming, and more!
Several months ago I saw this image and wondered if the plot shown in the original probelm was actually the solution to the stated PDE. In addition to being a fun problem it’s also a good excuse to...
Problem Evaluate [I=\int_0^1\ln(\Gamma(x))\,\mathrm dx.] Solution Let $x\mapsto1-x$ so [I=\int_0^1\ln(\Gamma(1-x))\,\mathrm dx.] If we take the average of these two integrals and apply the re...
Problem Evaluate [\int_0^1\left(\frac{1}{1-x}+\frac{1}{\ln(x)}\right)\mathrm dx.] Solution First we show that the integral [\displaystyle{I=\int_0^\infty\left(\frac{1}{e^x-1}-\frac{e^{-x}}{x}\...
The $n\text{th}$ harmonic number is given by [H_n=\sum_{k=1}^n\frac{1}{k}] and it is a well known fact that $H_n\to\infty$ as $n\to\infty$. Often this is proven by comparing the sum to the integr...
There are many variants of the “You should be able to solve this.” meme, the most famous being the Haruhi problem. Many have definite solutions but the problems are often complex or intractable, wh...
In general, proving the transcendence of a number is incredibly difficult, but there are some values that are amenable to rather quick transcendence proofs. Here we will show that [\alpha=\frac{\l...
Problem Evaluate [-\int_0^\infty\frac{\ln(\cos^2(x))}{x^2}\,\mathrm dx.] Solution Let [I=\int_0^\infty\frac{\ln(\cos^2(x))}{x^2}\,\mathrm dx] so [\begin{align}I &=\sum_{n=0}^\infty\int_{...
The double sum [\sum_{a=1}^{\infty}\sum_{b=1}^{\infty} \frac{a^2b^2}{\sinh(\pi(a+b))}(-1)^{a+b}] was asked about on math.se. Originally I tried evaluating this sum in a way that might make use of...
Problem Evalaute [\int_1^\infty\frac{\left{x\right}-\frac{1}{2}}{x}\,\mathrm dx] where the curly braces denote the fractional part of $x$. Solution First we may rewrite the integral as a serie...
Problem Evaluate [I=\int_0^{\pi/2}e^{-\sec^2(x)}\,\mathrm dx.] Solution Generalize the integral to [I(\alpha)=\int_0^{\pi/2}e^{-\alpha\sec^2(x)}\,\mathrm dx.] Using the substitution $x=\text{...
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