robjohn

West Hills, CA

none

Age: 55

the mean square The Mean Square
(with one standard deviation and several unusual ones)

aka Rob Johnson

$\LaTeX$ support for Chat

35m
comment Evaluating $\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt$
@Anastasiya-Romanova: I've added something similar to my answer, using a well-known result proven in another answer, to show the same thing, but bypassing the derivative and integration. Great idea, though! (+1)
1h
comment Evaluating $\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt$
@Anastasiya-Romanova: as I try to explain in $(4)$, $$\int_d^{d^c}\frac{t^{b/c-1}}{1-t}\mathrm{d}t$$ is between $d^{b/c-1}$ and $d^{b-c}$ times $$\int_d^{d^c}\frac1{1-t}\mathrm{d}t=\log\left(\frac{1-d}{1-d^{c}}\right)$$ and both $d^{b/c-1}\to1$ and $d^{b-c}\to1$ and $$\lim_{d\to1^-}\log\left(\frac{1-d}{1-d^{c}}\right)=-\log(c)$$
1h
comment About output of a function of Magma
If you edit the question, now that it has been undeleted, it will be placed on the reopen review queue. Of course, it only needs one more reopen vote.
1h
comment About output of a function of Magma
Yeah, it's a decent question, so I upvoted it, too. It is safe whether my vote counts differently with the script or not.
1h
comment About output of a function of Magma
I will have to check, but I thought that my all-mighty, unilateral vote might not get undone by Community.
1h
comment About output of a function of Magma
@DanielFischer: it had 2 votes, so I undeleted it. Thanks.
1h
comment Evaluating $\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt$
After the substitution, this becomes $$\begin{align}&\lim_{d\to1^-}\left[\int_0^d\frac{t^{a-1}}{1-t}\mathrm{d}t - \int_0^{d^c}\frac{t^{b/c-1}}{1-t}\mathrm{d}t\right]\\ &=\lim_{d\to1^-}\left[\int_0^d\frac{t^{a-1}-t^{b/c-1}}{1-t}\mathrm{d}t - \int_d^{d^c}\frac{t^{b/c-1}}{1-t}\mathrm{d}t\right]\\ &=\int_0^1\frac{t^{a-1}-t^{b/c-1}}{1-t}\mathrm{d}t - \lim_{d\to1^-}\int_d^{d^c}\frac{t^{b/c-1}}{1-t}\mathrm{d}t\\ &=\int_0^1\frac{t^{a-1}-t^{b/c-1}}{1-t}\mathrm{d}t + \log(c)\end{align}$$
1h
comment Evaluating $\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt$
@Anastasiya-Romanova: To perform the substitution $t\mapsto t^{1/c}$ on the second term in the integrand, we cannot simply break up the integral as $$\int_0^1\frac{t^{a-1}}{1-t}\mathrm{d}t-\int_0^1\frac{c\,t^{b-1}}{1-t^c}\mathrm{d}t$$ since neither of those integrals converge. Thus, we need to compute $$\lim_{d\to1^-}\left[\int_0^d\frac{t^{a-1}}{1-t}\mathrm{d}t-\int_0^d\frac{c\,t^{b-1}}{1-t^c}\mathrm{d}t\right]$$
2h
answered About output of a function of Magma
2h
revised Using Fourier Series to compute sums
fix typo
2h
revised Evaluating $\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt$
add another way to compute the integral
3h
awarded Guru
3h
awarded Necromancer
7h
comment Slice of pizza with no crust
@Venus: I used Mathematica to make the animation.
10h
awarded Revival
11h
revised Using Fourier Series to compute sums
fix typo
11h
comment Using Fourier Series to compute sums
@Semiclassical: 1) $\beta(2n)$ doesn't have any known closed form either. 2) In this answer, I analogously develop a recursion for $\zeta(2n)$. Is that close?
11h
answered Using Fourier Series to compute sums
15h
comment Using Fourier Series to compute sums
@Semiclassical: I derive a recursion for $\beta(2n+1)$ in this answer. $\beta(2n)$ is a harder nut to crack, like $\zeta(2n+1)$.
17h
revised Evaluating $\int_0^1 \frac{t^{a-1}}{1-t}-\frac{ct^{b-1}}{1-t^c}\ dt$
add explanation
1 2 3 4 5