Bit field in C; bytes and bits
Why isn't sizeof for a struct equal to the sum of sizeof of each member?
I'm trying to compute this integral, $$\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx \hbox{ where } 0&lt;a&lt;1$$ I drew a typical Pacman contour with branch cut at positive real axis. Then, we have $$\int_{\Gamma}\frac{\ln z}{z^{a}(z+1)}dz=\left(\int_{L_{1}}+\int_{L_{2}}+\int_{C_{R}}+\int_{C_{\epsilon}}\right)\frac{\ln z}{z^{a}(z+1)}dz$$ The function has a pole at $z=-1$. Then, by residue theorem, the whole integral becomes  \begin{align*} \int_{\Gamma}\frac{\ln z}{z^{a}(z+1)}dz&amp;=2\pi i\text{res}(\frac{\ln z}{z^{a}(z+1)};-1)\\&amp;=2\pi i\lim_{z\rightarrow-1}(z+1)\frac{\ln z}{z^{a}(z+1)}\\&amp;=2\pi i\lim_{z\rightarrow-1}\frac{\ln z}{z^{a}}\\&amp;=2\pi i\frac{\ln e^{i\pi}}{e^{i\pi a}}\\&amp;=2\pi i\frac{\ln1+i\pi}{e^{i\pi a}}\\&amp;=-\frac{2\pi^{2}}{(e^{i\pi})^{a}} \end{align*} On $L_1$, $z=xe^{i\epsilon}$ where $\epsilon\leq x\leq R$  On $L_{2}$, $z=xe^{i(2\pi-\epsilon)}$ where $\epsilon\leq x\leq R$  On $C_{R}$, $z=Re^{i\theta}$ where $\epsilon\leq\theta\leq2\pi-\epsilon$  Lastly on On $C_{\epsilon}$, $z=\epsilon e^{i\theta}$ where $\epsilon\leq\theta\leq2\pi-\epsilon$  So I tried to compute them separately and I anticipate the curves will both go to $0$ by ML inequality, and two linear integrals will consist of the term $\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx$ and hence simply solve for it. But I'm having trouble physically solving it out. I think the log is throwing me off. Can anybody please help me? Closing Reject Suggested Edit dialogue freezes review process Single service account running all SQL Servers Is there any negative performance impact for using the same service account and agent account to run SQL Server and SQL Agent respectively for all SQL server instances running at a small company of say 35 servers? The biggest database is 0.5 GB. any suggestions are welcome. Thank you. I've ran out of chocolate bars and didn't realize I could enchant my giant spoon and knight's armor. Is there any way to get more reliably? Make Chrome always execute with flags, for eg., directly opening an HTML file I want to run Chrome with a given set of flags, always  I created a shortcut and added the flag to it in the following way, so my target of the shortcut says:  C:\&lt;application-path&gt;\chrome.exe --user-data-dir=D:\Chrome   But when I double click an HTML file, or if any other application triggers Chrome, it looses the flag value. The result - A new chrome window opens with Default profile directory.  I want to make the flag permanent, so that its active irrespective of however Chrome executes.  I tried changing the following registry key:  HKEY_CLASSES_ROOT\ChromeHTML\shell\open\command   It was this earlier:  "C:\&lt;application-path&gt;\chrome.exe" -- "%1"    I changed to following attempts which did not work:  1. "C:\&lt;application-path&gt;\chrome.exe" -- "%1" --user-data-dir=D:\Chrome 2. "C:\&lt;application-path&gt;\chrome.exe" -- "%1" -- "user-data-dir=D:\Chrome" 3. "C:\&lt;application-path&gt;\chrome.exe" -- "%1 user-data-dir=D:\Chrome"   Conclusively, when I try to open a link from any application, it doesn't open in my current Chrome as a new tab, rather opens a new Chrome window with no plugins and extensions since its from other directory.  Why I have changed my default directory? - because the OS drive was getting low on memory due to heave Chrome caches and apps. As a physics student, I have occasionally run across the gamma function  $$\Gamma(n) \equiv \int_0^{\infty}t^{n-1}e^{-t} \textrm{d}t = (n-1)!$$  when we want to generalize the concept of a factorial.  Why not define the gamma function so that   $$\Gamma(n) = n!$$  instead?  I realize either definition is equally good, but if someone were going to ask me to choose one, I would choose the second option.  Are there some areas of mathematics where the accepted definition looks more natural?  Are there some formulas that work out more cleanly with the accepted definition? File and directory comparison tool? I'm looking for something that will compare directories and files like Beyond Compare does for windows.  Its mainly for use of source control, but I need to use it for deployment as well. I've been trying to wrap my head around the basic concepts of category theory, and I thought I would attempt to illustrate what I understand with the category of sets, probably the easiest example. Particularly, I've been trying to prove that $id_A$ (the identity morphism on $A$, for all $A \in Obj(\mathbf{Set})$) is $1_A \colon A \rightarrow A, x \mapsto x$.  This is a very intuitive and reasonable statement, and it's trivial to prove that $1_A$ is indeed an identity morphism on $A$, and I suppose uniqueness of $id_A$ can be demonstrated analogously to uniqueness of the identity element in a monoid (considering the subcategory which has $A$ as its only object, and endofunctions on $A$ as its only morphisms).  In this manner, it is not hard to prove that the proposition in the title is true, but this demonstration requires to make an assumption or guess as to what could $id_A$ be. Specifically, the scheme of the proof is: assume $id_A$ = $1_A$, see that it works with the definition of an identity morphism, show that the identity morphism is unique, and, in conclusion, $id_A$ can only be $1_A$. What I'm looking for, nonetheless, is a somehow more direct proof, that doesn't assume $id_A$ = $1_A$ at the start. I want to place myself in a state of little or no knowledge about sets and functions, and under this assumption, why would I assume $id_A$ = $1_A$ at first? Why not try with $id_\mathbb{Z}$ = $f \colon \mathbb{Z} \to \mathbb{Z}, x \mapsto x^2 + 1$, for example? It wouldn't work, but I don't have any reason to think that $1_\mathbb{Z}$ is a better guess for $id_\mathbb{Z}$.  I suppose that the proof for which I'm asking would work for categories of sets with additional structure, and probably for posets as well, although I'm not clear as to what modifications it would require to work.  Thanks. If photon energies are continuous and atomic energy levels are discrete, how can atoms absorb photons? If photon energies are continuous and atomic energy levels are discrete, how can atoms absorb photons? The probability of a photon having just the right amount of energy for an atomic transition is $0$. I've just got my first follower, and I'm wondering if I can get a second person to follow me. Is the game limited to one or two follower(s) or does it allow for a large number of followers? It really seems handy to me to have some followers protect me when I draw attention from a larger group, or when I try to steal something where I really shouldn't... I'm having trouble solving a Fitch Style Proof and I was hoping someone would be able to help me.  Premises: p ∧ q  Goal: q ∨ r  Would be an assumption (p ∧ q => q ∨ r) in first step a correct one?   Then, q ∨ r Implication Elimination: 2, 1  Why is it not correct?  Thank you very much for any help! bash if not multiple conditions without subshell? I want to combine multiple conditions in a shell if statement, and negate the combination.  I have the following working code for a simple combination of conditions:  if [ -f file1 ] &amp;&amp; [ -f file2 ] &amp;&amp; [ -f file3 ] ; then   # do stuff with the files fi   This works fine.  If I want to negate it, I can use the following working code:  if ! ( [ -f file1 ] &amp;&amp; [ -f file2 ] &amp;&amp; [ -f file3 ] ) ; then   echo "Error: You done goofed."   exit 1 fi # do stuff with the files   This also works as expected.  However, it occurs to me that I don't know what the parentheses are actually doing there.  I want to use them just for grouping, but is it actually spawning a subshell?  (How can I tell?)  If so, is there a way to group the conditions without spawning a subshell? How to get the list of IP addresses in a LAN? Probability of asymmetric random walk returning to the origin Consider the random walk $S_n$ given by $ S_{n+1} =  \left\{   \begin{array}{l}     S_n+2 &amp; \text{with probability $p$}\\     S_n - 1 &amp; \text{with probability $1-p$}   \end{array} \right. \ $  Assume that $S_0 = n &gt;0 $. What is the probability of eventually reaching the the point $0$?  Seems that the probability for this must be less than one; since it is not symmetric, it is not guaranteed to reach the origin if $p&gt;0.5$. It also seems like the positive starting value not being at the origin matters too, but I'm not sure how to compute the probability of eventually reaching the origin. [[ 0*10%300 ]] works on AIX 6.1 but not on AIX 7.1 (ksh) What are post-installation triggers?  These are executed when upgrading software. "Those who qualify will be awarded a certificate" or "those who qualify will be awarded certificates"? Can the Flash do things like build muscle faster than normal (do his muscles grow right after a work-out or meal)? Can he heal faster than normal? Does he have a faster immune system to stop all illness or at least make the flu last 2 minutes or something along those lines? Prove: $\lim\limits_{n \to \infty} \int_{0}^{\sqrt n}(1-\frac{x^2}{n})^ndx=\int_{0}^{\infty} e^{-x^2}dx$ I'd like your help with the following claim to prove:  $$\lim_{n \to \infty} \int_{0}^{\sqrt n}\left(1-\frac{x^2}{n}\right)^ndx=\int_{0}^{\infty} e^{-x^2}dx.$$  I think I should use the claim:     Let $a,b$ two real numbers and $\{f_n\}$ a sequence of continuous functions on $\left[a,b\right]$  which converges uniformly to $f$ on $[a,b]$. Then    $$\lim_{n\to\infty}\int_a^bf_n(t)dt=\int_a^bf(t)dt.$$   But for this, I must prove that $(1-\frac{x^2}{n})^n$ uniformly converges to $e^{-x^2}$.  How do I do that? One of the hardest and trickiest things to do is to prove this uniformly converges.. every function has its own way. I believe that the more examples I'll see the easiest it will be. Thanks again! Point Release upgrade from 14.04.1 to 14.04.2? I wanted to upgrade to 14.04.2 point release from 14.04.  What I did was:  sudo apt-get update &amp;&amp; sudo apt-get upgrade &amp;&amp;  sudo apt-get dist-upgrade -y   Then I wanted to upgrade to 3.16 kernel, since it wasn't done with the the above command, so what I did was:  sudo apt-get install linux-generic-lts-utopic xserver-xorg-lts-utopic libegl1-mesa-drivers-lts-utopic xserver-xorg-video-all-lts-utopic xserver-xorg-input-all-lts-utopic sudo apt-get install linux-signed-generic-lts-utopic   Now when I do:   uname -a I get   Linux Y500 3.16.0-33-generic #44~14.04.1-Ubuntu SMP Fri Mar 13 10:33:29 UTC 2015 x86_64 x86_64 x86_64 GNU/Linux  cat /etc/os-release | grep VERSION= I get  VERSION="14.04.2 LTS, Trusty Tahr"   And when I do lsb_release -a I get  Distributor ID: Ubuntu Description:    Ubuntu 14.04.2 LTS Release:    14.04 Codename:   trusty   I know that uname -a and lsb_release -a both refer to different info, but why does  cat /etc/os-release | grep VERSION= show different version?