I was trying to evaluate $ {\int_0^{\infty}\frac{\cos (x)}{1+x^2}}\text{d}x$ by Feynman's trick  It has been , but I have a specific question at one of the steps while doing it. (Not a duplicate) (My question is about why I get 2 different results when solving it in 2 different methods in a particular step.)  So I defined $I(t)= \int_0^{\infty}\frac{\cos (xt)}{1+x^2}\text{d}x$  and arrived at $I(t)= I''(t)$  so this means   $I(t) = \alpha e^t + \beta e^{-t}$  Method 1:  $I(0) = \alpha  + \beta $  and from the integral $I(0) = \int_0^{\infty}\frac{1}{1+x^2}\text{d}x$ = $\frac{\pi}{2}$  so $\alpha  + \beta = \pi/2 \tag1 $ and $I'(t) = \alpha e^t -\beta e^{-t}$, so $I'(0) = \alpha  -\beta $  from the integral $I'(t)= \int_0^{\infty}\frac{-x \sin (xt)}{1+x^2}\text{d}x$   so $I'(0)= 0 $  $\alpha - \beta = 0\tag2$  By (1) and (2) we get $$\boxed{I(t) = \frac{\pi}{4}e^t + \frac{\pi}{4} e^{-t}}$$    Method 2:  $I(0) = \alpha  + \beta $  and from the integral $I(0) = \int_0^{\infty}\frac{1}{1+x^2}\text{d}x$ = $\frac{\pi}{2}$  so $\alpha  + \beta = \pi/2 \tag3 $  $I(t)= \int_0^{\infty}\frac{\cos (xt)}{1+x^2}\text{d}x$  using integration by parts, $$u=\frac{1}{{{x}^{2}}+1}\quad,\quad dv=\cos tx$$ we have $$I(t)=\frac{\sin (t x)}{t({{x}^{2}}+1)}\left| \begin{matrix}    \infty   \\    0   \\ \end{matrix} \right.+\frac{2}{t}\int_{0 }^{+\infty }{\frac{x\sin (t x)}{{{({{x}^{2}}+1)}^{2}}}}\,dx $$ as a result $$I(t )=\int_{0}^{\infty }{\frac{2}{t} \frac{x\sin tx}{{{({{x}^{2}}+1)}^{2}}}\,}dx \,$$  As t tends to infinity we can see I(t) will tend to 0, which should mean $\alpha$ is 0, or else it will exponentially increase.  So now we get $$\boxed{I(t) = \frac{\pi}{2e^t}}$$    What's going wrong? Both seem valid and correct. This seeming loophole has been annoying me for a long time
Suppose that we do not know anything about the complex analysis (numbers). In this case, how to calculate the following integral in closed form? $$\int_0^\infty\frac{\cos x}{1+x^2}\,\mathrm{d}x$$
My questions is very simple. How can I set security setting for a PDF file created with PDFLatex?  I should set some password or print permissions, etc. Why 'Mrs.' isn't read as 'mistress'? Wasn't Mrs. short for mistress? Why do we read Mrs. as missus(or sometimes missis) instead of mistress? Finite $G$ with involutory automorphism $\alpha$ with no nontrivial fixed points. Proving properties of $\alpha$. The question at hand is:     Let G be a finite group and $\alpha$ an involutory automorphism of G, which doesn't fixate any element aside from the trivial one.    1) Prove that $ g \mapsto g^{-1}g^{\alpha} $ is an injection    2) Prove that $\alpha$ maps every element to its inverse    3) Prove that G is abelian    I think I've found 1).  I assume $ g_1^{-1}g_1^{\alpha} = g_2^{-1}g_2^{\alpha} $ and from this I get $ (g_2g_1^{-1})^\alpha = g_2g_1^{-1} $, so $g_2 = g_1$ (is this correct?)   For 2) I'm sort of stumped though, not sure how to start proving that, so any help please? Let $r$ be the smallest positive integer such that $a^r \equiv 1 \bmod{n}$. Prove $r \mid \phi(n)$. Let $a \in \mathbb {Z}$, $n &gt; 1$ a natural number with $\gcd(a, n) = 1$, and let $r$ be the smallest positive integer such that $a^r \equiv 1 \bmod{n}$. Prove that $r \mid \phi(n)$.  Euler's generalization of Fermat's Little Theorem: If $n \in \mathbb {N}$ and $m \in \mathbb{Z}$ such that $\gcd(m, n) = 1$, then $m^{\phi(n)} \equiv 1\bmod(n)$.  By Euler's generalization I would have $a^{\phi(n)} \equiv 1\bmod{n}$.  Then I could say $a^r \equiv a^{\phi(n)} \equiv 1 \bmod{n}$. At this point I can see that $r$ clearly must be some multiple of $\phi(n)$.  If I want to show that $r \mid \phi(n)$, then I can show that $\phi(n) = rk$, for some $k \in \mathbb {Z}$.  However, I am a bit stuck about how to make this last step showing that  $\phi(n) = rk$, where $k \in \mathbb {Z}$. How do you convert a numerical number to an Excel column name in C# without using automation getting the value directly from Excel.  Excel 2007 has a possible range of 1 to 16384, which is the number of columns that it supports. The resulting values should be in the form of excel column names, e.g. A, AA, AAA etc. Can black holes actually merge? If time stops at the event horizon, can we ever detect two black holes merging? In other words, if you are a short distance away, would you encounter a spherically symmetric gravitational field, or a dipole field? Getting Joker from the existing groups in Rummikub I have a group of tiles made up of 3 fours and a joker:     4 4 4 Joker   Now, if a player removes the joker, do they have to replace it with another 4? How to arrive at $\int_{0}^{1} f^3(x) \, dx < (\int_{0}^{1} f(x) \, dx)^2$? Define $f : [0,1] \to \mathbb{R} $ differentiable satisfying $f(0)=0$ and $0&lt;f'(x)&lt;1$. Prove:  $\displaystyle\int_{0}^{1} f^3(x) \, dx &lt; \left(\displaystyle\int_{0}^{1} f(x) \, dx\right)^2$ I found the following answer,  Define $F(x)=\displaystyle\int_{0}^{x} f^3(t) \, dt - \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)^2$ We have $$F'(x)=f^3(x)-2 \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)f(x)=f(x)\left[f^2(x)-2 \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)\right]$$ and $$\left[f^2(x)-2 \left(\displaystyle\int_{0}^{x} f(t) \, dt\right)\right]'=2f(x)\left[f'(x)-1\right]$$ from which we can arrive at the conclusion.  But I felt a little confused. In the past I usually proved the inequality by classical inequalities and something like Taylor expansion to use the condition about derivative. But this answer is little out of my expectation. Do we have other proofs or ideas to arrive at $\displaystyle\int_{0}^{1} f^3(x) \, dx &lt; \left(\displaystyle\int_{0}^{1} f(x) \, dx\right)^2$? And other thoughts about solving these types of issues? I think when we talk about $\left(\displaystyle\int_{0}^{1} f(x) \, dx\right)^2 &gt;\ldots \,$ it's a little difficult to deal with. I don't quite know what tools to use. Globally disable "include online search results" Updating .class file in jar If $f$ is an entire function where every power series expansion has at least one 0 term, show it is a polynomial Suppose $f$ is entire and that in every power series $f(z)=\sum_{n=0}^\infty c_n(z-a)^n$ at least one coefficient is $0$. Prove that $f$ is a polynomial. What's the best way to call a generic method when the type parameter isn't known at compile time, but instead is obtained dynamically at runtime?  Consider the following sample code - inside the Example() method, what's the most concise way to invoke GenericMethod&lt;T&gt;() using the Type stored in the myType variable?  public class Sample {     public void Example(string typeName)     {         Type myType = FindType(typeName);          // What goes here to call GenericMethod&lt;T&gt;()?         GenericMethod&lt;myType&gt;(); // This doesn't work          // What changes to call StaticMethod&lt;T&gt;()?         Sample.StaticMethod&lt;myType&gt;(); // This also doesn't work     }      public void GenericMethod&lt;T&gt;()     {         // ...     }      public static void StaticMethod&lt;T&gt;()     {         //...     } } How can I test the quality of a RNG? On electronics.stackexchange we had  about constructing a True Random Number Generator. Since the method relies on noise, which isn't deterministic, the only way to test the quality of the RNG seems to be empirical. Not being a statistician I suggested to test a long bit sequence for normality, but I have no idea what outcome is acceptable and what isn't. For instance counting single bits I guess we have a thumbs-up for a 499500/500500 distribution, but when is the ratio too much skewed to be acceptable? Same for 2-bit and longer sequences. Of course, if testing for normality is a Bad Idea™, I'd like to hear it, including better alternatives.    edit Diehard was mentioned a few times, but I'm not sure this answers my question. The normality test should give a normal distribution, with better bell curve approximations for longer sequences. Diehard seems also to have tests which either should result in normal or exponential distributions. But my question remains: how do I judge the results? Just by looking at the curve and discern a bell curve in it? To get back at my first example, a 499500/500500 distribution is definitely OK, and a 950000/50000 distribution definitely is a no-no, so where does the switch-over happen? Suppose $\sqrt6 = \frac pq$ where $p$ and $q$ have no common factors.  $$6 = \frac {p^2}{q^2}$$  $$6p^2 = q^2$$  So $q^2$ and therefore $q$ is divisible by $6$.   $$p^2 = \frac {q^2}{6}$$  So $p^2$ and therefore $p$ is divisible by $6$.   S0, $p$ and $q$ have a common factor $6$. Contradiction.  Therefore $\sqrt6$ is irrational.  Does the proof work? What are the main reasons the elves did not show up in force to defeat Sauron? During The Great Battle of the Last Alliance, there were an estimated 100,000 elves that fought with the armies of Isildur. This was one of the first scenes in the first LOTR movie. However, in the trilogy, it appears that only a small number of elves show up to Helms Deep and overall left the majority of the fighting to the humans. Why did the elves not participate during the rest of the war? I need to display attachments on a visual force page for the Case object.  This is the code I am using. I had it working for awhile, but it somehow broke down. It gets all the attachments, but it will not show the images.   &lt;apex:repeat value="{!Case.attachments}" var="att"&gt;             &lt;apex:panelgrid columns="1"&gt;                 &lt;h1&gt;Image/h1&gt;                 &lt;apex:image url="/servlet/servlet.FileDownload?file={!att.Id}"/&gt;             &lt;/apex:panelgrid&gt; &lt;/apex:repeat&gt; Multivariate differentiability verification I have tried an attempt on the following:  Let $F:U\in \Bbb{R}^n\to \Bbb{R}$ and $f:\Bbb{R}\to \Bbb{R}$ where $f$ is an even function. Now $F(\mathbf{x})=f(|\mathbf{x}|)$, where $|\ . |$ is the Euclidean norm. Given $f$ is $C^r$, I have to show that $F$ is $C^r$.  My attempt was to argument by induction. I showed that $F$ is $C^1$, basically using univariate chain rule on $\frac{\partial f\left(\sqrt{x_1^2+\dots+x_n^2}\right)}{\partial x_1}$ and arguing that it would be continuous using the fact that $f$ is $C^r$. Since the argument works for all $x_i$, it establishes first-differntiability for $F$ and continuity of the operator $DF(\mathbf{x})$. Then I moved on to show $F$ is $C^r$ assuming $F$ is $C^{r-1}$ and $f$ is $C^r$. Now what I did was partially differentiate:  $$ {\partial \over \partial x_2} \left[\frac{\partial ^{k-1}f}{\partial x_1^{k-1}}\right] = \underbrace{{\partial \over \partial |\mathbf{x}|} \left[\frac{\partial ^{k-1}f}{\partial x_1^{k-1}}\right]}_{(1)} . \underbrace{{\partial |\mathbf{x}| \over \partial x_2}}_{(2)} $$  My argument runs as follows: this partial derivative is continuous since $f$ is $C^r$ and hence $(1)$ is continuous. Continuity of $(2)$ is establishd in a similar manner, since $\sqrt{\mathbf{x}}$ is $C^r$ if $\mathbf{x} \neq \mathbf{0}$. Then the product would be continuous. I am uneasy on multivariate arguments and dont know if my approach is completely off track. I didnt use the even function property, so I am sure I might me missing something. Any hints and references would be welcome. Thanks! Setting page numbering on Word I have a word document. The first four pages are title and index. After that, the “Chapter 1” starts. I want page numbers to begin from the “Chapter 1” onwards only. The first 4 pages should not have any page number.  How do I achieve this? Laptop power supplies, does current matter? I have two laptops (same manufacture), with the same type of power connector.  However, the power supplies/transformers are slightly different.    The output on the first laptop's power supply is 15.6&nbsp;V at 8.0&nbsp;A. The output on the second laptop's power supply is 15.6&nbsp;V at 5&nbsp;A.  Clearly the voltages are the same, but the currents are different. I assume the second laptop's power supply can not be used on the first, because it can't supply enough power to the laptop.   However, can the first laptop's power supply be safely used on the second laptop? &quot;I hate wizards. If he so much as points a finger at me, you stick him, Hook.&quot; &quot;Remind me to find out the name of that flooring company so I can be sure never to buy so much as a carpet tile from it.&quot; &quot;I didn’t listen so much as survive the conversation.&quot;