$$\iint\limits_{Q} \frac{dx\,dy}{x^{-1}+|\ln y| - 1} \le 1 ,~~ Q = [0; 1] \times [0;1]$$
97452
Please, help me to prove that:  $$ \int_Q  \int  \frac{dxdy}{x^{-1} + |\ln y| - 1} \leq 1,$$ where   $$ Q =  [0; 1]  \times  [0; 1] $$ Any ideas how to start. Thank you.
7761
Compute  $$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$$
27106
The entire problem statement is:  Let $n&gt;1$ and let $U=B(0,1)\subset\mathbb{R}^n$. Show that $u:U\to\mathbb{R}$ given by  $$u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$$ is in $W^{1,n}(U).$  My attempt at the proof is as follows:  To show that $u\in W^{1,n}(U)$ it suffices to show that $|Du|\in L^{n}(U)$. Consider $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}|x|^{-2}\frac{x_i}{|x|},$$ which simplifies to, $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{x_i}{|x|^3}.$$ Thus, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{1}{|x|^2}$$ which can be manipulated to, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+|x|}\frac{1}{|x|}.$$  Moreover, since $U=B(0,1)$ we can bound the second term from above which gives, $$|Du|\leq\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}$$  Thus, we can instead show that $$\int_U\left(\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}\right)^n&lt;\infty.$$  Converting this into polar coordinates we have with $r=|x|$, $$\int_0^R\int_{\partial B(0,r)}\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\,dSdr=\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\int_{\partial B(0,r)}dSdr$$ Note that $\int_{\partial B(0,r)}\,dS=n\alpha(n)r^{n-1}$, which is the surface area of $B(0,r)$. Thus, we have then $$n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^nr^{n-1}\,dr=n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\right)^n\frac{1}{r}\,dr$$  And so this is where I get stuck in the problem, since I don't how I can evaluate that last integral to show it's finite.  Thank you for any help or feedback!
71517
I'm looking for several ways to prove that $$\int_{0}^{\infty }\sin(x)x^mdx=\cos\left(\frac{\pi m}{2}\right)\Gamma (m+1)$$  for $-2&lt; Re(m)&lt; 0$
109422
How to find $x\in L_{2}(0,\infty)$ \ $L_{q}(0,\infty)$  ,  $q\neq 2 $  i tried $\frac{1}{t.lnt}$ with various degrees on $t$ and $ln(t)$  Could you please help me with this question
74550
$$\lim_{x \to \infty} \left(1 +  \frac{a}{x}\right)^{x} = e^{a}$$     $1$. Without L´Hopital´s rule  $2$. With L´Hopital´s rule
87077
Suppose $\mu(X)&lt;\infty$. Show that if $\ f\in\ L^q(X,\mu)$ for some $1\le q&lt;\infty$ then $\ f\in\ L^r(X,\mu)$ for $1\le r&lt;q$   In the solution the assistant wrote this:  If $1\le r&lt;q$ then $q/r\ge 1$. Let p be the dual exponent to $q/r$ (i.e. $\frac{1}{q/r}+\frac{1}{p}=1)$  then by Hölder Inequality  $\int_{X}|f|^rd\mu\le\underbrace{(\int_{X}|f|^qd\mu)^{r/q}}_{??????}(\int_X1^pd\mu)^{1/p}&lt;\infty$  for me the middle part doesn't make sense (it is not the hölder inequality) can somebody show it ?  instead of this can't we say (only in this case) that   $\int_{X}|f|^rd\mu\le (\int_{X}|f|d\mu)^r$ then  $(\int_{X}|f|d\mu)^r\le ((\int_{X}|f|^qd\mu)^{1/q}(\int_X1^pd\mu)^{1/p})^r&lt;\infty$  ?
87316
Let $f(x_1,x_2)=\frac{1}{(x_1+ix_2)^2} \chi_{ \{ (-1,1) \times (-1,1) \}}$  Part A: Evaluate the $L^p(R^2)$ norm of $f(x_1,x_2)\chi_{ \{ (x_1,x_2): x_1^2+x_2^2 \leq 1 \}}$ for every $1 \leq p \leq \infty$ (this function is =f inside the disk of radius 1 and is =0 outside the disk)  $||f||_{L^p(R^2)}=(\int\int_{\{ (x_1,x_2): x_1^2+x_2^2 \leq 1 \}} |f(x_1,x_2)|^p dx_1dx_2)^{1/p} =(\int\int_{\{ (x_1,x_2): x_1^2+x_2^2 \leq 1 \}} (\sqrt{x_1^2+x_2^2})^{-2p} dx_1dx_2)^{1/p} =(\int_0^{2 \pi}\int_0^1 r^{-2p} rdr d \theta)^{1/p} =(\int_0^{2 \pi}d \theta \times \int_0^1 r^{-2p+1}dr)^{1/p} =(2 \pi \times r^{-2p+2}]_0^1)^{1/p} =(2 \pi \times \frac{1}{2-2p})^{1/p} =(\frac{\pi}{1-p})^{1/p}$  Please check that the work is correct.  Part B: Evaluate $L^2(R^2)$ norm of $f(x_1,x_2)$  $||f||_{L^2(R^2)}=(\int\int_{\{ (-1,1) \times (-1,1) \}} |f(x_1,x_2)|^2 dx_1dx_2)^{1/2} =(\int_{-1}^!\int_{-1}^1 (\sqrt{x_1^2+x_2^2})^{-4} dx_1dx_2)^{1/2} =(\int_{-1}^!\int_{-1}^1 \frac{dx_1dx_2}{(x_1^2+x_2^2)^2} dx_1dx_2)^{1/2}$  If I continue down this path I end up using multiple trig and u substitutions, how can I use Part A to solve this problem?
118596
This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.  $$\lim_{(x,y) \to (0,0), x+y \neq 0}{\frac{\ln(1-x-y)}{x+y} } $$  My attempt:   Let $\xi = -x-y $. Then $\xi \to 0$ whenever $(x,y) \to (0,0)$ and $x+y \neq 0 \iff \xi \neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):  $$\lim_{\xi \to 0, \xi \neq 0}{\frac{\ln(1+\xi )}{-\xi}}$$   If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?
87229
Is there a closed form expression for $\text{n}^{\text{th}}$ coefficient of Maclaurin series of  $$\ln(1-x+x^2)$$
25938
$\dfrac{n}{2} \le \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^n - 1} \le n $  I've Tried for hours but didn't got any striking idea. I don't have any efforts to show rather than induction. Please Try If you 've non-inductive proof.
13721
$$\dfrac{1}{2}\ln(x+7)-(2 \ln x+3 \ln y)$$  Our professor let's us solve this but i do not understand how $\ln$ works. He says it has same properties with $\log$ but i still don't get it. What's the difference of the two?
105538
$$\sum_{n=2}^{\infty} \frac{ (-1)^n} { \ln(n) +\cos(n)}$$
67772
I'm given that $f \in L^1$ and $f(x) \neq 0$ a.e. and trying to prove that for q > 0  $$\lim_{q \to 0} (\int_0^1 |f|^q)^{1/q} = e^{\int_0^1 \ln|f|}$$  My work:  I took the log of both sides to reduce the problem to  $$\lim_{q \to 0} \frac{1}{q} \ln(\int_0^1 |f|^q) = \int_0^1 \ln|f|$$  Since $\ln$ is a concave function I used Jensen's inequality to show that  $$\frac{1}{q} \ln(\int_0^1 |f|^q) \geqslant \int_0^1 \frac{1}{q}\ln|f|^q = \int_0^1 \ln|f| $$  So if the limit exists, I know  $$\lim_{q \to 0} \frac{1}{q} \ln(\int_0^1 |f|^q) \geqslant \int_0^1 \ln|f|$$  Now I can't make any more progress to reverse the inequality.  Maybe I need a completely different approach.  Also what about the necessity of $f \in L^q$? Please help.
99286
The limit is    $$\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$$  The problem is I don't know if I can calculate it normally like with a change of variables or not. Keep in mind that I'm not allowed to use L'Hôpital's rule nor the $\mathcal O$-notation.
59910
Any help would​ be appreciated.  My initial attempts were using Lagrange's Mean Value Theorem first between $f(0)$ and $f(1)$ to show that $f'(t)=1$ for some $t$ in $(0,1)$.  Now applying LMV between $f'(t)$ and $f'(0)$, and $f'(t)$ and $f'(1)$ I could prove it if t lies in either $(0,1/4]$ or in $[3/4,1)$. Don't really know where to go next.
115924
Prove that $$\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}$$