15,507 reputation
11351
bio website math.stackexchange.com/…
location
age
visits member for 11 months
seen 15 mins ago

Joshua King came to Cambridge from Hawkshead Grammar School. It was soon evident that the school had produced someone of importance. He became Senior Wrangler, and his reputation in Cambridge was immense. It was believed that nothing less than a Second Newton had appeared. They expected his work as a mathematician to make an epoch in the science. At an early age he became President of Queens’; later, he was Lucasian Professor. He published nothing; in fact, he did no mathematical work. But as long as he kept his health, he was an active and prominent figure in Cambridge, and he maintained his enormous reputation. When he died, it was felt that the memory of such an extraordinary man should not be permitted to die out, and that his papers should be published. So his papers were examined, and nothing whatever worth publishing was found.


6h
comment What fraction of sunlight could we deliberately block out with an occulter?
What an occult question. :-)
7h
comment Proper way to set up “Pac-Man” contour integral
Alternately, let $t=\dfrac1{1+x}$, and recognize the expression of the beta function in the new integral. Then, by expressing it in terms of the $\Gamma$ function, and using Euler's reflection formula, rewrite it in terms of elementary trigonometric functions.
17h
comment Contour Integral: $\int^{1}_{0}\frac{1}{\sqrt[n]{1-x^n}}dx$
See beta function.
17h
comment How to apply the alternating series test to the series $\sum (-1)^{n+1} n/2^n$?
This is a polylogarithm.
19h
answered Evaluating $\int x^2 \sqrt{x^2-1} dx$
22h
comment How to simplify this summation
This is a polylogarithm.
22h
answered How to simplify this summation
22h
revised How to simplify this summation
edited body
1d
comment Sum of series with two binomial coefficients
Hint: The key concept here is binomial series. On one hand, $\displaystyle\sum_{k=0}^\infty{k+n\choose n}x^k~$ is the binomial expansion of $\dfrac1{(1-x)^{n+1}}$ . On the other hand, $\displaystyle\sum_{k=n}^\infty{k\choose n}x^k=\dfrac{x^n}{(1-x)^{n+1}}$ .
1d
revised Where to learn integration techniques?
added 75 characters in body
2d
answered Evaluate $\int_0^\infty\frac{dl}{(r^2+l^2)^{\frac32}}$
2d
comment Show that $\int^\infty_0\left(\frac{\ln(1+x)} x\right)^2dx$ converge.
Its value is $\dfrac{\pi^2}3=2~\zeta(2)$, and, in general, by generalizing the value of the integrand's exponent to n, the definite integral can be expressed as a sum of zeta functions, $I_n=\displaystyle\sum_{k=2}^na_k~\zeta(k)$, where $a_k\in$ Q.
Jul
27
comment Prove ${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{2/3}}=\frac{3^{3/2}}{2^{4/3}5^{5/6}\pi }\Gamma^3\left(\frac13\right)$
Perhaps it would be better to ask directly for a general approach to evaluating $\displaystyle\int_{0\text{ or }-1}^1(x^n+a)^m(1-x^p)^q~dx$. For $a=0$, the connection to the beta function is obvious.
Jul
27
comment Definite integral $\int_{-\pi/2}^{\pi/2}\cos^{2}\left(\theta\right)\,{\rm d}{\theta} $
See Wallis' integrals.
Jul
27
comment Find the flaw in my 1-page proof of the Four Color Theorem
See Alfred Kempe and Four Color Theorem: History.
Jul
27
comment Help with Complex integration
A similarly interesting identity would be $~\displaystyle\int_0^\infty\frac{\sin x}{\sinh x}~=~\frac\pi2\tanh\frac\pi2$
Jul
27
comment Prove $_2F_1\left(\frac13,\frac13;\frac56;-27\right)\stackrel{\color{#808080}?}=\frac47$
I used $100-200$ decimals of precision, and a polynomial degree of $20$.
Jul
27
comment Prove $_2F_1\left(\frac13,\frac13;\frac56;-27\right)\stackrel{\color{#808080}?}=\frac47$
Other special values are $x=-\dfrac13$ and $x=+\dfrac12$, for which the minimal polynomials are $9x^3-8$ and $16x^6-27$.
Jul
27
comment Prove $_2F_1\left(\frac13,\frac13;\frac56;-27\right)\stackrel{\color{#808080}?}=\frac47$
By changing the argument from $-27$ to $5$, we get one of the roots of $~5^5x^6+12^3=0$.
Jul
26
comment Prove ${\large\int}_0^\infty\left({_2F_1}\left(\frac16,\frac12;\frac13;-x\right)\right)^{12}dx\stackrel{\color{#808080}?}=\frac{80663}{153090}$
So, you are denying that $\displaystyle\sum_{n=1}^\infty n=-\frac1{12}$ , or that $\displaystyle\sum_{n=1}^\infty1=-\frac12$ ? :-)