A Difficult Integral

I'm wanting to keep up my LaTeX skills so I'm returning to an integral, specifically: $$  \int\limits_0^1 \, \frac{x}{x^3+1}\text{d}x$$The integral is not all that difficult really but it does involve a fair number of steps to get it right. Here is a graph of what is being attempted:

I used symbolab to work through the solution which turns out to be:\[\frac{1}{9}(\sqrt{3}\pi-\ln{(8)})\]The first step in the solution is to factorise the denominator. This is easy because \(x^3+1 \) is a sum of two cubes and so can be written as \((x+1)(x^2-x+1)\). Using partial fractions, the integral can be converted into the following form:\[ \frac{1}{3} \int\limits_0^1 \frac{x+1}{x^2-x+1} \text{d}x-\frac{1}{3} \int\limits_0^1 \frac{1}{x+1} \text{d}x \]The RHS integral is straightforward but the LHS one poses more difficulties. The way forward with the former is to write the quadratic in the denominator as a sum of two squares:\[ \frac{1}{3} \int\limits_0^1  \frac{x+1}{(x-\frac{1}{2})^2+\frac{3}{4}} \text{d}x \]This newly changed denominator can then be divided into the two parts of the numerator to yield:\[ \frac{1}{3} \int\limits_0^1 \frac{x}{(x-\frac{1}{2})^2+\frac{3}{4}}+\frac{1}{3} \int\limits_0^1 \frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}} \]I'm not going to go into the details here but, with the appropriate substitutions, the LHS of this integral can be integrated to a logarithm and the RHS to an arctan using the formulae:\[ \int \frac{2t}{t^2+1} \text{d}t = \ln|t^2+1| + C \text{ and }\int \frac{1}{t^2+1} \text{d}t=\tan \!^{-1}{t}+C \]The indefinite integral for \( x/(x^3+1) \) turns out to be: \[ \frac{1}{6} \Big( \ln|4x^2-4x+4|+2\sqrt{3} \, \tan^{-1} \big( \frac{2}{\sqrt{3}}(x-\frac{1}{2}) \big) \Big)-\frac{1}{3}\ln|x+1|+C \]The definite integral for the limits of 0 and 1 is what was quoted earlier at the beginning of this post. In conclusion, I must say that Blogger is still inconsistently handling the display mode for LaTeX: sometimes it renders the LaTeX using the Javascript and other times it returns an image of the rendered code. Annoying.

Gaps between Twin Primes

Today I turned 24917 days old. This number is prime and forms a twin prime with 24919. One claim that 24917 has to fame is that it marks the end of a gap of 496 between consecutive twin primes. After 24421 (the larger of twin pairs), there is a gap of 496 numbers until 24917 (the smaller of twin primes).

Source

At the right is a list of the record breaks. Of course, I won't live to experience the next record gap of 628 days between 62299 and 62927. However, I won't have to wait long until the next twin prime pair: 24917 and 24919. There is a gap of only 58 days between 24919 and 24977.

Of course, there's a lot of interest as to whether twin primes go on forever. It's very likely of course but it hasn't been proven. Progress has been made recently however, by Yitang Zhang, James Maynard and Terence Tao with his Polymath Project. The breakthrough relates specifically to the gaps between prime numbers rather than the gaps between successive pairs of twin prime numbers. There's a good account of these developments in this article from quantamagazine.

In any case, I felt it was important to celebrate the end of this latest record gap as my endless, numbered days unravel.