The question is self-explanatory, why does the integral of  $$x^x$$does not exist? Can someone give a proof or reason?
I'm trying to find $\int x^x \, dx$, but the only thing I know how to do is this:  Let $u=x^x$.  $$\begin{align} \int x^x \, dx&amp;=\int u \, du\\[6pt] &amp;=\frac{u^2}{2}\\[6pt] &amp;=\dfrac{\left(x^x\right)^2}{2}\\[6pt] &amp;=\frac{x^{2x}}{2} \end{align}$$  But it's certain that this isn't the correct way to evaluate that, and the answer must be wrong.
Show that no application $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, of $C^k$ class, $k \geq 1$ can be injective How can I proof this:     Show that no application $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, of $C^k$ class, $k \geq 1$ can be injective, i.e., there are $A,B \in \mathbb{R}^2$ such that $A \neq B$ and $f(A) = f(B)$ (Hint: show that given $(a,b) \in \mathbb{R}^2$ in every open neighbourhood of $(a,b)$ there are points $(c,d),(c_1,d_1)$ where ${\partial}_1f(c,d) \neq 0$ or ${\partial}_1f(c_1,d_1) \neq 0$).   The problem is that I have no idea how to use the hint.  Thanks for all your comments. Given three variables, y and x, which are positive continuous, and z, which is categorical, I have two candidate models given by:  fit.me &lt;- lmer( y ~ 1 + x + ( 1 + x | factor(z) ) )   and  fit.fe &lt;- lm( y ~ 1 + x )   I hope to compare these models to determine which model is more appropriate. It seems to me that in some sense fit.fe is nested within fit.me. Typically, when this general scenario holds, a chi-squared test can be performed. In R, we can perform this test with the following command,  anova(fit.fe,fit.me)   When both models contain random-effects (generated by lmer from the lme4 package), the anova() command works fine. Owing to boundary parameters, it is normally advisable to test the resulting Chi-Square statistic via simulation, nonetheless, we can still use the statistic in the simulation procedure.  When both models contain only fixed-effects, this approach---and, the associated anova() command---work fine.  However, when one model contains random effects and the reduced model contains only fixed-effects, as in the above scenario, the anova() command doesn't work.  More specifically, I get the following error:   &gt; anova(fit.fe, fit.me)  Error: $ operator not defined for this S4 class   Is there anything wrong with using the Chi-Square approach from above (with simulation)? Or is this simply a problem of anova() not knowing how to deal with linear models generated by different functions?  In other words, would it be appropriate to manually generate the Chi-Square statistic derived from the models? If so, what are the appropriate degrees of freedom for comparing these models? By my reckoning:  $$ F = \frac{\left((SSE_{reduced}-SSE_{full})/(p-k)\right)}{\left((SSE_{full})/(n-p-1)\right)} \sim F_{p-k,n-p-1} $$  We are estimating two parameters in the fixed effects model (slope and intercept) and two more parameters (variance parameters for the random slope and random intercept) in the mixed-effects model. Typically, the intercept parameter isn't counted in the degrees of freedom computation, so that implies that $k=1$ and $p=k+2=3$; having said that I'm not sure if the variance parameters for the random-effects parameters should be included in the degrees of freedom computation; the variance estimates for fixed-effect parameters are not considered, but I believe that to be because the parameter estimates for fixed effects are assumed to be unknown constants whilst they are considered to be unknowable random variables for mixed effects. I would appreciate some assistance on this issue.  Finally, does anybody have a more appropriate (R-based) solution to comparing these models? Fake induction proofs I have a series of layers of lines (shapefiles) in My Project. The CRS of some are different and to merge them they all must have same CRS.  When I use Processing/Toolbox/Qgis_algorithims/Reproject_layer the reprojection only works if it is allowed to save to temporary file somewhere (it will not change the CRS when I reproject if I try to save and replace it in My Project directory);   I must remove the original from the layer list and reproject the temp file (without changing the CRS) to get it into my project and name it appropriately.  Is there a better way to reproject a layer in Qgis? centered logo ok? I'm designing a website for a rather unknown software company. The company likes the design but cannnot get along with our desicion to center their logo on the top of their page, although all the other elements are centered. Therefore a left aligned logo wouldn't look good. I know, it seems all UX studies prove them right, but positioning the logo in the center isn't a unpopular practice anymore.   Furthermore I think the way we read a website has adapted the last 10 years. Are there any studies about this topic that could help prove my point?  Thanks What do viruses do? There are those big green spiked things that my friends call viruses. I don't know what they do and I want to know. Do the viruses do something to the device you are playing on? Are they good or bad? Any javascript event occurring when user clicks Stop load button? Is there a way to transfer more than one Pokemon at a time? Is there a way to transfer more than one Pokemon at a time to the professor?  Or possibly an addon that gives me this ability?  Vendoring all the trash Pokemon for candies individually is really annoying and time consuming. Prove that for all sets A and B, A ∩ B = ∅ implies ( A ∪ B ) - B = A In the next proof we avail ourselves of the next lemma: For all sets A and B, ( A ∪ B ) - B = A. Proof: Let A and B be arbitrary sets and let x ∈ ( A ∪ B ) - B. x ∈ ( A ∪ B ) - B ⇔ x ∈ ( A ∪ B ) ∧ x ∉ B ⇔ ( x ∈ A ∨ x ∈ B ) ∧ x ∉ B ⇔ ( x ∈ A ∧ x ∉ B ) ∨ ( x ∈ B ∨ x ∉ B) For p ∧ ~ p ≡ F and p ∧ q ⇒ p where p and q are prepositions, ( x ∈ A ∧ x ∉ B ) ∨ ( x ∈ B ∨ x ∉ B ) ⇔ x ∈ A ∨ F By modus tollendo ponens, x ∈ A ∨ F ⇔ x ∈ A So then, x ∈ ( A ∪ B ) - B ⇔ x ∈ A And ( A ∪ B ) - B = A Therefore, the statement for all sets A and B, A ∩ B = ∅  implies ( A ∪ B ) - B = A is true trivially. Is this proof right? I'm automating a web application (the Mantis bug tracker) and I'm getting an interesting response header from it, called Refresh:  HTTP/1.x 200 OK ... Refresh: 0;url=my_view_page.php   It seems to be acting the same way that  does, and the meta refresh technique implies that it is an equivalent of a header in HTTP.  Problem is, I can't find any mention of the Refresh header in the  or any other definitive documentation on how it should be parsed and what the browser should do when it encounters it.  What's going on here? Log4Net: Programmatically specify multiple loggers (with multiple file appenders) How to (programmatically, without xml config) configure multiple loggers with Log4Net? I need them to write to different files. How can I change the class of an HTML element in response to an onclick or any other events using JavaScript? Acer Aspire E15 will not dual boot I installed Ubuntu 14.04 onto my Acer Aspire E15. The PC has a 500 Gigabyte HDD.   During the installation it divided the HDD into separate partitions. However every time I turn on the PC it goes directly into Windows 8.1. It doesn't allow you the opportunity to select which OS you want to use.   Question 1: How do I set the PC so it will allow me to select the OS?  Question 2: And if #1 isn't possible. How to I remove Ubuntu from the PC and recover the partition for Windows Usage of "they" / "them" / "their" when the person's gender is not known How can I change the page URL without refreshing the page? Clocks for multiple timezones in GNOME Shell I looked everywhere but I couldn't find a way to display multiple clocks in GNOME 3. In classic GNOME and Unity, it is possible. Also, for some strange reason, additional timezones cannot be added to the clock using the time-date applet in classic GNOME. Any solutions? I am trying to send an email using a Visualforce template and I want that template to include an icon. Is this possible? What's the difference between the list methods append() and extend()? MySQL convert date string to Unix timestamp How prove there is no continuous functions $f:[0,1]\to \mathbb R$, such that $f(x)+f(x^2)=x$. Prove that there is no continuous functions $f:[0,1]\to R$, such that   $$ f(x)+f(x^2)=x. $$   My try. Assume that there is a continuous function with this property. Thus, for any $n\ge 1$ and all $x\in [0,1]$, \begin{align*} f(x)&amp;=x-f(x^2)=x-\big(x^2-f(x^4)\big)=x-x^2+\big(x^4-f(x^8)\big)=\cdots\\ &amp;=x-x^2+x^4-\cdots+(-1)^n\left(x^{2^n}-f\big(x^{2^{n+1}}\big)\right) \end{align*}  since $f(0)=0$ and $\displaystyle\lim_{n\to \infty}x^{2^{n+1}}=0$ for any $x\in(0,1)$,it follows by the continuity of $f$ that $\displaystyle\lim_{n\to \infty}f\big(x^{2^{n+1}}\big)=0$, hence $$f(x)=x-x^2+x^4-x^8+\cdots+(-1)^nx^{2^n}+\cdots$$ for any $x\in (0,1)$  Why do I have prove that there is exists such a continuous functions $f$? Maybe  my example is wrong? Why?  if my   method is wrong,then How prove this problem ? Thank you.