Integrating $\int_{0}^{\infty} \frac{(\log x)^2}{x^2+x+1}$ using residue theorem
Definite integral, quotient of logarithm and polynomial: $I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x$
What is the easiest way to update/upgrade an offline Ubuntu machine with the help of a Windows machine that has internet access?  This question also applies to installing new packages offline and not only updates/upgrades.  The Linux system shall remain completely cut off from the Windows machine, meaning there should be no ethernet connection or the likes between the two computers.  Update:  Please have patience, since I haven't updated this since two weeks. There are some complications getting apt-offline to work properly on Lubuntu and all other methods seem to be too old and deprecated:  AptOff - last update 2006 AptMedium - last update 2007 Keryx - last update 2011, looks like this project was discontinued  I hope I will be able to somehow update the offline machine, finally.  While I was still not able to update/upgrade the offline system due to authentication errors, I accepted an answer, because it's taking too long. What is the easiest way to match non-ASCII characters in a regex? I would like to match all words individually in an input string, but the language may not be English, so I will need to match things like ü, ö, ß, and ñ. Also, this is in Javascript/jQuery, so any solution will need to apply to that. Objects with transparent texture not always show transparency in 3dview For some reason objects don't show their transparency all the time.  In this image the circles in red are suposed to be transparent like the ones in green.    When I select the object in the back, all objects in front of it become opaque.    When I select the two objects that are opaque, then they become transparent as they sould be.    The display is in GLSL mode  These are the material and texture properties      I'm not worried about the final render, just the editing view. I would like to see the actual transparency at editing time only, because the scene is to be imported in a third party game engine.  Blender is v2.68   is the blend file Induction on a sum The left hand side has terms involving  $\binom{n}{m}= \dfrac{n!}{(n-m)!m!}$    $$1+\dfrac{1}{2}\binom{n}{1} +\frac{1}{3}\binom{n}{2}+...........+\frac{1}{n+1}\binom{n}{n} = \dfrac{2^{n+1}-1}{n+1}$$  I've done induction and proved $P(0)$ holds and also $P(1),P(2)$ holds (Just in case)  Now I've found $P(K+1)$ so the sum on the left is going to equal $\dfrac{2^{n+1}-1}{n+1}$ + $\dfrac{1}{n+2}$  and the right hand side is going to equal $\dfrac{2^{n+2}-1}{n+2}$  My problem is coming with the algebra though. I can't get those sides to equal. I can't get rid of the $n+1$ in the denominator.   Anyone have any ideas for me? They'd be much appreciated!  P.S. Someone gave me a hint and said that you can use the binomial theorem to solve this. Why do I get "Exception; must be caught or declared to be thrown" when I try to compile my Java code? If $|a| = 12, |b| = 22$ and $\langle a \rangle\cap \langle b\rangle \ne e$, prove that $a^6 = b^{11}$ $ d_1,d_2\mid n\iff {\rm lcm}(d_1,d_2)\mid n\ $ [LCM Universal Property] I know this is a very common corollary of the class equation. And I know how to do it by using class equation.   But can you do it bu using group action, maybe find a nice set for G to act on? How do you convert a byte array to a hexadecimal string, and vice versa? How can you convert a byte array to a hexadecimal string, and vice versa? SQL Server 2012 Page Life Expectancy resets to 0 after about 50 days I've noticed an odd behavior on a 2-server HA cluster and I was hoping someone could confirm my suspicion, or maybe offer some other explanation... Here is my setup:   A 2-server SQL 2012 SP1 installation SQL AlwaysOn HA has been enabled for a few databases CPUs are 2.4GHz, 4 cores RAM is 34 GB (it is an AWS instance, hence the odd number) Resource utilization is relatively low - each server has 14+ GB memory free, and SQL is not capped on how much memory to use Disk access time is fine - rarely going over 15ms/Read or Write Databases are not big - 1 GB, 1.5 GB, 7.5 GB SQL server process is using 16 GB Private Bytes, 15 GB Working Set   Overall, no resource issues are noted. Now for the odd part. SQL is not restarted (process has been running for almost 6 months) but it seems that every ~50 days, the Page Life Expectancy counter drops to (almost) 0. Up until that point it climbs steadily, no drops. Here is a perf graph:    When I look at the counter data (I don't have the exact number, just an hourly aggregation) it seems the PLE counter value reached about 4,295,000 sec (roughly 50 days) every time (at least every time I have data for).   My crazy theory is that the PLE number is held as milliseconds as an unsigned long int (which has a limit of 4,294,967,295) and at 49.71 days it resets, either by design, or because of a bug. This would explain the behavior of the two servers and the identical pattern they have. Or it could be something totally different and I am just not making any sense. :)  Has anyone seen anything like that, or can explain this behavior?  P.S. I saw  post, but my case seems slightly different.  P.P.S. This is a repost - I originally posted it , but was advised the audience here is more appropriate.  Thanks! Memory and CPU optimizers: Are they gimmicks? Capacitance of a single charged plate? Lets say we have a single plate that has a charge of $+Q$ on it. A plate with charge $-Q$ is infinite distance away. Will the plate with $+Q$ have a capacitance associated with it? Why or why not? I was thinking that because the distance between the plates is infinity, the capacitance is zero, but according to my TA, that is not the case. There is always some sort of "absolute" capacitance for the plate with $+Q$ (a capacitance will exist between the plate and air?). Can someone help me out in understanding this? In Inno Setup is it possible to add conditional statements to what in the [Run] section gets executed? Alternatively is it possible to manually update the built in progress bar? Basically I have 2 MSIs included and using Inno Setup as a bootstrapper, and depending on user input one or both of the MSIs are to be installed.  I have something working using Exec statements in CurStepChanged but it doesn't update the progress bar as the files are extracted and it looks like the installer is stalled.  I guess the end result is I want some progress bar updates while the files are extracted to the temp folder.  The following is my current code:  procedure CurStepChanged(CurStep: TSetupStep); var     ResultCode: Integer; begin     if(CurStep = ssInstall) then begin         if(InstallServer) then begin             ExtractTemporaryFile('ServerSetup.msi');             Exec('msiexec',ExpandConstant('/i "{tmp}\ServerSetup.msi" /qb INSTALLDIR="{code:GetInstallPath}\Server\" ALLUSERS=2'),'', SW_SHOW, ewWaitUntilTerminated, ResultCode);         end;         if(InstallClient) then begin             ExtractTemporaryFile('ClientSetup.msi');             Exec('msiexec',ExpandConstant('/i "{tmp}\ClientSetup.msi" /qb INSTALLDIR="{code:GetInstallPath}\Client\" ALLUSERS=2'),'', SW_SHOW, ewWaitUntilTerminated, ResultCode);         end;     end; end; Import a python module without the .py extension How to move an element into another element? I'm getting an error - didnt understand relationship when running this query in dev console. I have set the child relationship the lookup field to IncEvents. The lookup field is on the MyRule object -   select Id, Name, (select Id from IncEvents_r) from Namespace_MyRule__c  I noticed that the editor is displaying a single underscore between the namespace and the object name but I did enter a double underscore. I have just bought a Nikon 50mm f/1.8D prime lens.  I took this picture (leg of my glasses frame on a mouse, and a LED desk lamp is on top of it) : f1.8, 1/80s, ISO 200   The purple fringing is right in the middle of the picture. Does it happen very often to this lens? Can Wubi use the ISO I already have? The Windows Installer for Ubuntu desktop instructions say that an ISO image for Ubuntu is downloaded during the installation process.  But what if I already have an ISO image for Ubuntu. Can I make the Windows Installer use that, so it doesn't have to download an ISO image again?  12.04 LTS and 12.10 are the versions of Ubuntu I'm interested in installing, in case that matters. The axiom of choice asserts that for any set $X$ there exists a function $f : (2^X − \{\emptyset\}) \rightarrow X$ such that for any nonempty $A \subseteq X$, $f(A) \in A$. Show that this is equivalent to the assertion that every relation includes a function with the same domain. I am given this problem in my assignment. My approach is following. I have a set $X$. Then it has a choice function $g : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$. Suppose $X$ has a relation $R \subseteq X \times X$. $dom(R) \subseteq X$. Then the required function $f : dom(R) \rightarrow X$ is $$f(x) = g(\{y\ |\ (x, y) \in R\})$$ Conversely, suppose $X \supseteq Y$ is a non-empty subset. Let $$R_Y = \{(x, y)\ |\ y \in Y, x \in Y, (p, q) \in R_Y \Rightarrow p = x\}$$ Then there is a function $f_Y : dom(R_Y) \rightarrow X$. But $dom(R_Y)$ is singleton from the definition of $R_Y$. Then $im(f_Y)$ is also singleton. We define our choice function as $f = \{(Y, y)\ |\ y \in im(f_Y)\}$ Are my arguments correct?  Edit.1: what I tried to do for converse is, for any subset $Y$, I construct a relation $Y_1 \times Y$ where $Y \supseteq Y_1 = \{y\}$. Then there is a function $f_Y : Y_1 \rightarrow Y$. Then if $f$ is my choice function I define $f(Y) = f_Y(y)$. May be I should define $R_Y$ as $$R_Y = \{(x, y)\ |\ y \in Y; x \in Y; (p, q), (p', q) \in R_Y \Rightarrow p = p'\}$$ I understand there may be multiple $R_y$'s. But can't I define a singleton set as $\{s\in S\ |\ a, b \in S \Rightarrow a = b\}$? This can be $\{s\}$ $\forall s \in S$. But isn't it a valid specification?  Edit.2: This is the final solution with the help from @Asaf.  If we assume axiom of choice then for any $X$ there is a choice function $f : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$ such that $f(A) \in A$. Suppose $R$ is a relation between $P$ and $Q$. Then we have a choice function $f_Q$ for $Q$. Let $f_R(p) = f_Q(\{q\ |\ pRq\})$. The function $f_R : P \rightarrow Q$ and $dom(f_R) = dom(R)$. So this is our required function.  Conversely, if every relation includes a function with same domain. Let $R$ be a relation between $\mathcal{P}(X) \setminus \left\{\emptyset\right\}$ and $X$ such that $aRb \Rightarrow b \in a$. Now $R$ includes a function $f : \mathcal{P}(X) \setminus \left\{\emptyset\right\} \rightarrow X$. Then $xRf(x) \Rightarrow f(x) \in x$. So $f$ is the choice function for $X$. As linguistics student, I have to put more than one thing on a character every now and then. For example, I need to type ā̆. I thought I could simply do that by typing \={\u{a}}, but then the macron is placed above an empty space in front of the character.  How can I solve this? Thanks  P.S. I'm using MiKTeX 2.9, running on Windows 8; in case it might be of use.