Evaluate $$I=\int_{0}^{1} \frac{\ln(1-x)\ln^2(1+x)\:dx}{x}$$  We have $$\frac{\ln(1-x)}{x}=-\sum_{k=1}^{\infty}\frac{x^{k-1}}{k}$$  Hence $$I=-\sum_{k=1}^{\infty}\left(\frac{1}{k}\int_{0}^{1}x^{k-1}\ln^2(x+1)\:dx\right)\tag{1}$$  Now first lets try to evaluate $$J=\int_{0}^{1}x^{k-1}\ln^2(1+x)\:dx$$ Using Integration By parts and using the fact that $\lim_{x \to 0}x{\ln{x}}=0$  $$J=\frac{\ln^2 2}{k}-\frac{2}{k}\int_{0}^{1}\frac{x^k\ln(1+x)\:dx}{1+x}=J-\frac{2}{k}W \tag{2}$$  Now $$W=\int_{0}^{1}\frac{x^k\ln(1+x)\:dx}{1+x}$$  Writing $$x^k=(1+x-1)^k=(1+x)^k-\binom{k}{1}(1+x)^{k-1}+\binom{k}{2}(1+x)^{k-2}+\cdots+(-1)^k$$  Hence $$W=\sum_{m=0}^{k}(-1)^m\binom{k}{m}\int_{0}^{1}(1+x)^{k-m-1}\ln(1+x)\:dx$$  Using the Fact that when $a\ne -1$: $$\int_{0}^{1}(1+x)^a\ln(1+x)\:dx=\frac{\ln 2 \times 2^{a+1}}{a+1}-\frac{2^{a+1}-1}{(a+1)^2}$$  and when $a=1$ $$\int_{0}^{1}\frac{\ln(1+x)}{1+x}\:dx=\frac{\ln^2 2}{2}$$  Then $$W=\left(\sum_{m=0}^{k-1}(-1)^m\binom{k}{m}\left(\frac{\ln 2 \times 2^{k-m}}{k-m}-\frac{2^{k-m}-1}{(k-m)^2}\right)\right)+\left((-1)^k\frac{\ln^2 2}{2}\right)$$  Using change of variable for the summation as $m=k-r$ we get:  $$W=(-1)^k\left(\sum_{r=1}^{k}(-1)^r\binom{k}{r}\left(\frac{\ln 2 \times 2^{r}}{r}-\frac{2^{r}-1}{(r)^2}\right)\right)+\left((-1)^k\frac{\ln^2 2}{2}\right)$$  Any way to proceed further?
Question: Integrate$$\int\limits_0^1dx\,\frac {\log(1-x)\log^2(1+x)}x=-\frac {\pi^4}{240}$$   I'm curious as to if there is a way to integrate this. I've tried using integration by parts to get$$I=-\frac {\pi^2}6\log^22+2\int\limits_0^1dx\,\frac {\operatorname{Li}_2(x)\log(1+x)}{1+x}$$However, I'm not sure how to continue even further. The polylog in the second integrand seems a bit intimidating and I don't see how the first term even helps.
What does a "real" quantum computer need for cryptanalysis and/or cryptographic attack purposes? The cryptographic world has been buzzing the word "quantum" for a while now (even the NSA is currently preparing itself for a post-quantum crypto world) and quantum-related hardware engineering is evolving constantly. For example: the 5-qubit quantum computer created at MIT by using the technique of ion traps succeeded in prime-factorizing 15. Does that mean that since it succesfully managed that, that it is a all-purpose quantum computer which could be used for cryptanalysis and/or cryptographic attacks?   If that's not the case, what exactly would a "real" quantum computer need (think: rough description of expected technical abilities, aka specs) to enable its users to use it for cryptanalysis and/or cryptographic attack purposes? And - ignoring rumours about potentially existing but confidential governmental projects - does any such system already exist today? Our DM keeps interrupting our long rests In the latest campaign, we're still around the third level, but every single night we've spend in the wild so far, we were interrupted by hard combat encounters (as in DMG-worded "hard"), forcing us to use our resources (like HP, spell slots, Ki, once-per-day stuff) and not restoring any of those lost the previous days. I feel like the , as it lasts four to six rounds, so we usually say that we'd like to resume our rest but the answer is always "you've been interrupted, your rest is over, you can't restart any until dusk". So the only real long rests we could have always were in town.  We've always used watch-turns, and being interrupted in the first night of the campaign, we now always spend one hour to find a safe place to rest and we make that intent very clear to the DM. But yet, kobolds, goblins, ghouls and gnolls always find us in the middle of the night, but never when we travel for ten hours during the day (which is comparable to "fast travel" in his own word). Once on the second day in the wild, we were faced to a deadly encounter (as in DMG-worded "deadly") which we survived only by fleeing. And we're kind of forced to get back to the deadly encounter, which is more than one day away from the nearest town.  We're five players and I'm the only one with PnP RPG experience... from ten years ago. I don't remember being this resourceless. We asked if it was a survival campaign and the answer was that it was only a normal introduction campaign found on the Internet that he spiced up a bit. We've also asked if we did anything wrong and the answer was that we followed the normal story.  The DM is mastering his second campaign (the first one is still ongoing, but with veteran players). The DM also plays a lot of survival campaigns as a player.  Is this situation normal? If yes, why? If not, what should I say to the DM?  Note that I'm fully aware of the "long rest" rule as written on page 186 of the PHB. This is why I encouraged my team to say that we resume our rest. (thanks Bloodcinder) How to make \left, \right pairs of delimiter work over multiple lines? Prove that $\frac{ 5^{125}-1}{ 5^{25}-1}$ is a composite number First, my understanding is that the long-term capital gains tax rate is 0% for those whose marginal rate on ordinary income is 10% or 15%, and (ignoring the highest 39.6% bracket) the rate is 15% for everyone else. (See  and .)  My interest is in what happens at the crucial break between the 15% and 25% bracket, which is $73,800 for married/joint returns in 2013 (and presumably will be higher in 2014, though I haven't been able to find out what that is or if it has even been set yet). Let's say a household has $64,000 in wage income, and sells stock for a $10,000 profit.   Does their income for the purposes of establishing their tax bracket go to $74,000? If so, is the entire $10,000 profit subject to the 15% long-term capital gains rate? Is any part of the wage income subject to the higher marginal tax rate?   Since the amount of income subject to the higher marginal rate is, well, marginal, part 3 seems somewhat less important than part 2. The real issue is whether the capital gains rate hits the entire profit, since, in that case, it seems the household should only sell enough stock to generate a $9000 profit, and sell the remainder in a future year. A function $f: \mathbb{R} \to \mathbb{R}$ is said to have the intermediate-value property if for any $a$, $b$ and $\lambda \in [f(a),f(b)]$ there is $x \in [a,b]$ such that $f(x)=\lambda$.  A function $f$ is injective if $f(x)=f(y) \Rightarrow x=y$.  Now it is the case that every injective function with the intermediate-value property is continuous. I can prove this using the following steps:   An injective function with the intermediate-value property must be monotonic. A monotonic function possesses left- and right-handed limits at each point. For a function with the intermediate-value property the left- and right-handed limits at $x$, if they exist, equal $f(x)$.   I am not really happy with this proof. Particularly I don't like having to invoke the intermediate-value property twice.  Can there be a shorter or more elegant proof? What could be the reason covid19 death data shows strong periodic trends? The covid19 death data for US shows strong periodic trends. It does not make sense to me that people die periodically.   What could be the reason to having such data? I thought it might be the reporting artifacts, for example, people always update data at Monday, but the period seems to be 3-4 days. What is the CVE-2014-6271 bash vulnerability (Shellshock) and how do I fix it? How to determine the architecture of an installation or Live CD/DVD? How can I determine which architecture (x86 or x64) my Ubuntu installation medium is for?  I don't have it installed, I just need to know which files on the installation media I can use to determine the architecture. consider $r&gt;0 , p&gt;1$ and $K \subset B(x_0 , 2r) \subset R^n$ . $K$ compact. Define the sets :  $$A = \{ u \in C^{\infty}_{0} (B(x_0 , 2r)); \textit{ such  that }  \  u=1 \textit{ in a open  neighborhood of K} \}$$  $$ B =  \{  u \in  H^{1,p}_{0} (B(x_0 , 2r)) \cap  C (B(x_0 , 2r))  : u=1 \  on \  K  ,  \  0 \leq u \leq 1\}$$  I want to show that   $$ \displaystyle\inf_{u \in A} \displaystyle\int_{B(x_0 , 2r)} |\nabla u|^p  \ dx= \displaystyle\inf_{u \in B} \displaystyle\int_{B(x_0 , 2r)} |\nabla u|^p   \ dx$$  For this i am trying to aproximate (in the sense of the  norm of $H^{1,p}$) the function of B , by function of $A$.  Somenone can give me a hint ?  thanks in advance Trouble Viewing Whole HDR Image as cycles world When I load an HDR image into an environment texture using Cycles Render, only a small fraction of the image appears. I have a feeling it's an easy fix, but I can not find what needs to be changed. Referencing bib files with spaces in the filename I am using a Mac with TeX Live and Mendeley for automatically generating .bib files, some of which have spaces in the file names. I would like to know how to reference such files from LaTeX as using "" or \ doesn't seem to work. How do I disable tabs for  tag I am using &lt;a&gt; tags for links on a web page. How do I disable the Tab key from selecting either of them? In music, when two or more pitches are played together at the same time, they form a chord. If each pitch has a corresponding wave frequency (a pure, or fundamental, tone), the pitches played together make a , which is obtained by simple addition. This wave is no longer a pure sinusoidal wave.  For example, when you play a low note and a high note on a piano, the resulting sound has a wave that is the mathematical sum of the waves of each note. The same is true for light: when you shine a 500nm wavelength (green light) and a 700nm wavelength (red light) at the same spot on a white surface, the reflection will be a superposition waveform that is the sum of green and red.  My question is about our perception of these combinations. When we hear a chord on a piano, we’re able to discern the pitches that comprise that chord. We’re able to “pick out” that there are two (or three, etc) notes in the chord, and some of us who are musically inclined are even able to sing back each note, and even name it. It could be said that we’re able to decompose a Fourier Series of sound.  But it seems we cannot do this with light. When you shine green and red light together, the reflection appears to be yellow, a “pure hue” of 600nm, rather than an overlay of red and green. We can’t “pick out” the individual colors that were combined. Why is this?  Why can’t we see two hues of light in the same way we’re able to hear two pitches of sound? Is this a characteristic of human psychology? Animal physiology? Or is this due to a fundamental characteristic of electromagnetism? How to trigger a check for updates in Firefox programatically or from a command line? Is there a command line switch for firefox.exe or an "about:" URL that will either force an update check or at least display the Help/About dialog, which checks for updates and tells if you're running the latest version?    One site claimed that the "about:" URL was the same as menu Help -> About, but it's not.  I built a program to automate the updating of various programs on my machine, and most programs have command line tools for checking for updates. Windows update has wuauclt.exe, Java has jucheck.exe. For some applications, I can even automate the interface, but it's difficult in Firefox, because the main window title is unpredictable (it depends on which web page is active), and all Firefox windows seem to use the exact same window class name. Proving bessel identity $[J_0(x)]^2 + 2[J_1(x)]^2 + 2[J_2(x)]^2 + \cdots = 1$ and $|J_0(x)|\le 1$ The generating function for a Bessel equation is:      $$g(x,t) = e^{(x/2)(t-1/t))}$$      Using the product $g(x,t)\cdot g(x,-t)$ show that:      a) $$[J_0(x)]^2 + 2[J_1(x)]^2 + 2[J_2(x)]^2 + \cdots = 1$$      and consequently:      b)      $$|J_0(x)|\le 1, \forall x$$      c) $$|J_n(x)| \le \frac{1}{\sqrt{2}}; n=1,2,3,\cdots$$   For a) I tried the product:  $$e^{(x/2)(t-1/t))}\cdot e^{(x/2)(-t+1/t))} = 1$$  I at least arrived at the right side of the equation. Since this generates the bessel functions, I should arrive at something related to $J_n$ in the left side. I know that  $$e^{(x/2)(t-1/t))} = \sum_{n=0\infty}^{\infty} J_n(x)t^n$$  $$e^{(x/2)(-t+1/t))} = \sum_{n=0\infty}^{\infty} J_n(x)(-t)^n$$  but it's not just a matter of multiplying coefficients from the two infinite series, right?  For $b$, I tried to use  but it's not as obvious  I have no idea how to deal with $c$, could somebody help me? I am confused with the grammar here. Which is grammatically correct?       At which time will you call me?     At what time will you call me?    And why? Real analysis derivate problem. How can I prove this? Let $f$ be differentiable in $[a,b]$ and $f(a)=0$. If $\exists M \in R$ such that $\vert f'(x) \vert \leq M \vert f(x)\vert \ $,  $\forall x \in [a,b]$, then $f(x)=0, \ \forall x \in [a,b]$.  $\\$ I have an idea, but I can´t see it clear. I have arrived to the inequality $\vert f(x)\vert \leq  K\vert f(y)\vert $ for $y \in (a,x)$, supposing $k&gt;0$ and repeating the process $\vert f(x)\vert \leq  K^n\vert f(z)\vert$ for $a&lt;z&lt;y&lt;x$. If $K&lt;1$, then if $n \rightarrow \infty \ $, $f(x)=0$. That´s a sort of intuitive idea. Theorem style with name as argument Is there some way to get a theorem type which will make a theorem with the name passed as a parameter? In other words, is there a way to define "named" such that   \begin{named}[NAME] text \end{named}   will print   "NAME: text"?  I tried using the definition \newtheorem*{named}{}, but that results in   "(NAME): text", which is quite different from what I am trying to do.  I know that I could also create a new instance of \newtheorem for each named theorem I want to use, but that is clunky and inconvenient. Is there a better way? Let $R$ be a commutative ring with $1$. Suppose that all ideals $I \neq R$ are prime. Prove that $R$ is a field.   Help me some hints.  Thanks a lot.