@Chris'ssis Okay... I wasn't sure how many coefficients were generalized. I could see either the constant or the linear term being generalized. I think both should have relatively nice forms.
@r9m this one is another proposal that was sent to AMM (everything is recorded in my e-mail, when I sent it to them)
@r9m ask Donald or whoever you want to find the result of this one and see what happens. The result there is possible due to my research, it was UNKNOWN. (I give you the link (again) when you're on)
I doubt you can find it in any paper around the world.
I solved almost all open problems from Ovidiu's book (that's because in some cases I only solved some particular cases).
Okay, let's call $$f_N(x) = \sum_{n=1}^N a_n \sin (nx),\quad f(x) = \sum_{n=1}^\infty a_n \sin (nx).$$ Then all $f_N$ are $2\pi$-periodic functions with mean $0$, and $f_N\to f$ uniformly. Let $A = \{ x \in [0,2\pi] : f(x) > 0\}$, $B = \{ x \in [0,2\pi] : f(x) < 0\}$. We can use an alternating series argument to see that $\int_0^\infty \frac{x}{1+x^2}f(x)\,dx$ exists as an improper Riemann integral. Since $\lvert e^{inz}\rvert \leqslant 1$ in the upper half-plane, $f$ is continuous in the closed
upper half-plane, and holomorphic in the upper half-plane. So we can use the residue theorem to evaluate the integral.
Hmm, slowly, we would need to see that $f(z)$ decays nicely as $\operatorname{Im} z \to +\infty$.
Yes, okay, no problem.
Err, Obviously I forgot to replace $\sin (nz)$ by $e^{inz}$ above. Imagine that done.
$$\int_0^{\pi/2} x \cot(x) \csc^2(x) \log(\sec(x))=?$$
I hope to avoid talking about my achievements, some will say that I definitely brag with them, and they will be right thinking like that since I talk too much about that these days even if I don't like to do it.
I got a question. If I prove something is no larger than $n^{\frac23+\epsilon}$, where $\epsilon$ is any positive constant, and $n\to \infty$ but is finite, will I be right to say I proved that thing is no larger than $n^{\frac23}$, as I can always choose smaller epsilon.. no, right?
@Daniel gah, so to prove what I want I need to prove it is certainly smaller than $n^{\frac23+\epsilon}$, and have $\epsilon$ as any positive, even dependent on $n$, right?
The question is whether I can conclude it is no larger than $\Theta(n^{2/3})$ by proving it is certainly smaller than $n^{2/3+\epsilon}$ if I choose the right definition for the epsilon - i.e. just positive
@BalarkaSen It's a perfectly fine Lebesgue integral even if the integrand is left undefined at $0$. But since one can continuously extend the integrand to $[0,1]$, even Riemann has no problems.
I need some starting ideas, hints for proving that
$$2 \le \int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \le 3$$
I already checked that with Mathematica that numerically says that
$$\int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \approx 2.577632915067858 $$
If one day you are faced with the moral dilemma of doing something morally right to help other people and yet this thing is illegal and you could be punished for doing it, what would you do?
@Chris'ssis Yes, I have been thinking of doing this thing (which is not well defined) for years, but I have not done it, because I am busy with other things like solving my own mental problems.
@Chris'ssis @Alizter The moral question I just asked you has bothered me for years. It is very painful. In the end, I will have to make a decision and act accordingly and face the consequences.
In this sick world, too many sick politicians make sick decisions which cause others suffering.
@JasperLoy Events happen, but some people will never be changed. I often see in the interviews I have very stupid people that have no hope. What can I do?
@JasperLoy someone (in an interview) was telling me he's a great mathematician but he didn't know how to compute $$\lim_{x\to 0}\frac{1-\cos(x)}{x^2}$$ without l'Hopital.
I need some starting ideas, hints for proving that
$$2 \le \int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \le 3$$
I already checked that with Mathematica that numerically says that
$$\int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \approx 2.577632915067858 $$
