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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.


3h
comment Definite integral involving powers and logarithm
Expand $\ln(1-x)$ into its Taylor series, switch the order of summation and integration, and let $t=x^b$, then employ the expression of the beta function.
3h
comment Finite integral with goniometric functions, $\int_0^{\infty} \frac{8\sin^4(\pi f t)\tan^2(\pi f/2)}{(\pi^4 \tau^2 f^3) }df$
Letting $x=\dfrac\pi2f$, and $a=2~\tau$, after extracting all constants outside the integral sign, we are finally left with evaluating $\displaystyle\int_0^\infty\frac{\sin^4(ax)\tan^2x}{x^3}dx$, which seems to yield nice solutions of the form $\ln A$ for $a\in2~\mathbb Z$, and $A\in\mathbb Q$, where the growth of A's numerator and denominator seems to sky-rocket, despite the fact that their ratio is only about $\dfrac a2$ .
6h
comment Prove ${\large\int}_0^\infty\frac{\ln x}{\sqrt{x}\ \sqrt{x+1}\ \sqrt{2x+1}}dx\stackrel?=\frac{\pi^{3/2}\,\ln2}{2^{3/2}\Gamma^2\left(\tfrac34\right)}$
You seem to have forgotten the minus sign, since the quantity is negative. Also, letting $x=\sinh^2t$, the integral can be rewritten as $\displaystyle\int_0^\infty\frac{\ln(\sinh t)}{\sqrt{\cosh(2t)}}dt$. Perhaps something similar exists in Gradshteyn-Ryzhik ?
7h
answered minimal polynomial given an algebraic number
8h
comment Convergence of ${\large\int}_{-\infty}^\infty J_0(x)\,J_0(x+a)\,dx$
Since $\dfrac\pi2$ appears in the various integral formulae for this and other related functions, $\dfrac2\pi$ was a first obvious choice. But both visual and numerical verifications, by integrating on intervals of the form $(-10^n,+10^n)$ , seemed to infirm that. So, going out on a limb, I tried $a=\sqrt{\dfrac2\pi}$ , which appeared to actually work.
8h
comment Convergence of ${\large\int}_{-\infty}^\infty J_0(x)\,J_0(x+a)\,dx$
I plotted the integrand's graphic for various values of the parameter a, and couldn't help but notice that it was centered for $a=0$, but shifted for other values. I logically assumed that this was due to its lack of symmetry, so I changed it to $J_0\bigg(x\pm\dfrac a2\bigg)$. Then, given its similarity with $\dfrac{\sin x}x$, I tried to guess the value of a for which the positive and negative parts cancel out each other's divergent tendencies. I also plotted the hyperbolae $y=\pm\dfrac1{2x}$ for visual aid, which implied that a had to be in between $\pm1$. $[\ldots]$
20h
comment Integral ${\large\int}_0^\infty\frac{\ln x}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}\ dx$
It can be rewritten as $\quad\displaystyle\int_0^\infty\frac{\ln(\sinh x)}{1+\sinh x}\cdot e^{x/2}~dx,\quad$ for which Mathematica is able to return a hideous expression.
21h
comment A closed form for $\int_{0}^{\pi/2} x^3 \ln^3(2 \cos x)\:\mathrm{d}x$
Search for a solution of the form $\quad\displaystyle\sum\alpha_k\cdot\zeta(k)\cdot\pi^{n-k},\quad$ where $\alpha_k\in\mathbb Q~$, and $~k\in\Big\{3,5,7\Big\}$.
23h
comment Prove $\left(\dfrac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$
$$x^x~=~a\qquad=>\qquad x~=~\dfrac{\ln a}{W(\ln a)}$$ See Lambert W function for more details. Now, for $a=\ln2$, we have $x=0.4000402600\ldots$
1d
awarded  Necromancer
1d
comment Gamma Function Removable Singularity
$\Gamma(x)\sim\dfrac1{x}-\gamma~$ for $~x\to0$, where $\gamma$ is the Euler-Mascheroni constant.
1d
comment Evaluate $ \int_{0}^{\frac{\pi}{2}}\frac{1}{(1+x^2)(1+\tan x)}\,\mathrm dx$
@achillehui: Coming from you, this means a lot. :-)
2d
comment Existence of improper integral
Its value is $\pi\ln2$.
2d
comment Evaluation of $ \int\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$
Hint: $x^4+1=(x^2+x\sqrt2+1)(x^2-x\sqrt2+1)$.
2d
comment Convergence of ${\large\int}_{-\infty}^\infty J_0(x)\,J_0(x+a)\,dx$
$$a=\sqrt{\frac2\pi}\qquad\overset{?}=>\qquad\int_{-\infty}^\infty J_0(x-a)\cdot J_0(x+a)~dx~\overset{?}=~1.$$
Jul
21
comment If $p$ and $q$ are positive real numbers, show that $\sum_{k=2}^\infty(-1)^k\frac{(lnk)^p}{k^q}$ converges
This series is nothing else than the p-th derivative of the Dirichlet $\eta$ function of argument q.
Jul
21
revised $\int_{0}^{\pi/2}\ln\left(1+4\sin^4 x\right)\mathrm{d}x$ and the golden ratio
added 3 characters in body
Jul
21
answered $\int_{0}^{\pi/2}\ln\left(1+4\sin^4 x\right)\mathrm{d}x$ and the golden ratio
Jul
21
comment $\int_{0}^{\pi/2}\ln\left(1+4\sin^4 x\right)\mathrm{d}x$ and the golden ratio
As long as $t>0$, the same principle applies. For $-1<t<0$, it's even easier.
Jul
21
comment $\int_{0}^{\pi/2}\ln\left(1+4\sin^4 x\right)\mathrm{d}x$ and the golden ratio
$1+4\sin^4x~=~(1-2i\sin^2x)(1+2i\sin^2x),\qquad\ln(ab)=\ln a+\ln b,\qquad\Re(\pm2i)>-1$.