Use the defintion of a logarithmic function to prove $$ \frac{1}{x+1} \le \ln(1+\frac{1}{x}) \le \frac{1}{x}$$ for any positive integer x.   ( Using defintion : $e^y=\ln^{-1}y$ )     Attempt:    Using the definition,  $e^y=\ln^{-1}y$ ,    I tried to prove the right hand side of the equality first by $ 1 +\frac{1}{x} \gt \frac{1}{x}$,     $\ln(1 +\frac{1}{x})\lt \frac{1}{x} \Rightarrow   e^{\ln(\frac{x+1}{x})} \le e^{\frac{1}{x}}$ but I am really stuck and lost on how I should do it so that one of the sides there is ln and the other side is the value and also for the equality to be $\le$ as $ 1 +\frac{1}{x} \gt \frac{1}{x}$  is strictly less than and is not equal.
121784
I looking for help with proving the following inequality. Any relevant logarithmic identities would be great. Tried differentiating and taking limits and I'm lost as to how to approach this.   $$\frac 1{x+1}&lt;\ln\left(\frac{x+1}x\right)&lt;0.5\left(\frac 1x+\frac 1{x+1}\right)$$
79323
I am trying to show that, for all $x&gt;0:$  $$\frac{1}{\sqrt{x}}\geq \frac{\ln x}{x-1}$$  This inequality is closer than I expected. I have tried exponentiating, power series, and have achieved nothing. I would really appreciate some help. Below is a graph of the two functions for small $x:$
35056
$$\int^1_0\frac{x^b-x^a}{\ln x} d x =\ln{{\frac{1+b}{1+a}}}$$ The  part inside $\ln$ is absolute value. A solution including integration under the integral can be found . Which used the identity $$\int^1_0x^{\alpha} dx =\frac{1}{1+\alpha}$$. Multiplies both side by $d{\alpha}$ and integrates with respect to $\alpha$. Are there other methods to prove this identity?  Thank You.
56810
Show that $$\int_{\ln\frac{3}{2}}^{\ln2}\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\,\mathrm dx=\ln\frac{9}{5}$$  How would you integrate this logarithmic function in order to obtain $\ln(\frac{9}{5})$.
72110
Prove that $\frac{1}{a+1}&lt;\ln \frac{a+1}{a}&lt;\frac{1}{a},a&gt;0$  First inequality using MVT:  $\frac{1}{a+1}&lt;\ln \frac{a+1}{a}:$  $f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$  $f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}&gt;f(1)\Rightarrow f(a)&gt;f(1)$  $\frac{1}{a+1}-\ln \frac{a+1}{a}-\frac{1-2\ln 2}{2}&gt;0$  This is not the starting inequality.   Is there something wrong in this method?
105432
Is there a way I can derive the value of the integral  $ \int_0^1 \ln(x)\ln(1-x)dx$  using the fact that  $\displaystyle\sum_{i=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$  ? (the actual value of the integral is $2-\frac{\pi^2}{6}$)  Thanks in advance
6034
We are dealing with $z \in \mathbb{C}$. I know that $$ \left(1+ \frac{z}{n} \right)^n \to e^{z} $$ as $n \to \infty$. So intuitively if $z_n \to z$ then we should have $$ \left(1+ \frac{z_n}{n} \right)^n \to e^{z}. $$ If $z_n \in \mathbb{R}$ I would be happy writing $$ \exp \left(n \log\left(1+ \frac{z_n}{n} \right) \right) = \exp \left(z_n \frac{\log\left(1+ \frac{z_n}{n} \right)-\log(1+0)}{\frac{z_n}{n}-0} \right) \to \exp(z \cdot 1) $$ where here we are using the definition of derivative. But if $z \in \mathbb{C}$ the logarithm is multivalued and I'm not sure that the same calculation is allowed. Is there a different way to show this for complex $z$?
101791
Let $f$ be an entire function with $|f(z)|\le 100\log|z|,\forall |z|\ge 2,f(i)=2i, \text{ Then} f(1)=?$  I have no idea how to solve this one!  $g(z)={f(z)\over \log|z|}$ Then Can I say $g$ is constant by Liouville Theorem?
117830
How to integrate $\frac{1}{x\log x}$?  Could you give me some ideas on how to integrate this? thanks.  i've tried setting $u=(\log x)^{-1}$.   $\dfrac{\mathrm du}{\mathrm dx} = x^{-1}$  But it didnt work...
7969
How to prove      $$I=\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx=4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2\ln2\ G$$   Where $ \operatorname{Li}_3(x)$ is the the trilogarithm function and $G$ is the Catalan constant.  Variant approaches are appreciated.
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.
42702
Of course, it is easy to see, that the integral (or the antiderivative) of $f(x) = 1/x$ is $\log(|x|)$ and of course for $\alpha\neq - 1$ the antiderivative of $f(x) = x^\alpha$ is $x^{\alpha+1}/(\alpha+1)$.  I was wondering if there is an intuitive (probably geometric) explanation why the case $\alpha=-1$ is so different and why the logarithm appears?  Some answers which I thought of but which are not convincing:   Taking the limit $\alpha=-1$ either from above or below lead to diverging functions. Some speciality of the case $\alpha=-1$ are that both asymptotes are non-integrable. However, the antidrivative is a local thing, and hence, shouldn't care about the behavior at infinity.
95777
I know that $\int \frac{1}{x}~dx = \ln|x| + C$ and I know the antiderivative method works for all powers of $x$ except $-1$. But why is that the case? I am still in high school and teachers aren't really helpful with these questions.   Edit: I do realize dividing by zero is meaningless. That's not the question. Just because dividing by zero is meaningless doesn't mean mathematicians just choose to ignore it and work just make up a new answer. There has to be an explanation, and that's what I am asking for. The explanation.
103492
Computing the integration.  $\int_0^1\frac{\ln x}{1-x}d x$  I have no idea ... how to do that?  ... The answer is $-\frac{\pi ^2}{6}$ I get it by software.
83811
I would like to prove the following inequality without integration; could you help?   $$\frac {1}{x+1}\leq \ln (1+x)- \ln (x) \leq \frac {1}{x}, \quad x &gt; 0. $$  I can however differentiate this.  Thanks in advance.
27738
Prove that      $$ \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx = \ln\left(\frac{a}{b}\right) $$   My Attempt:  Define the function $I(a,b)$ as  $$ I(a,b) = \int_{0}^{\infty}\frac{e^{-bx}-e^{-ax}}{x}\,dx $$  Differentiate both side with respect to $a$ to get  $$ \begin{align} \frac{dI(a,b)}{da} &amp;= \int_{0}^{\infty}\frac{0-e^{-ax}(-x)}{x}\,dx\\ &amp;= \int_{0}^{\infty}e^{-ax}\,dx\\ &amp;= -\frac{1}{a}(0-1)\\ &amp;= \frac{1}{a} \end{align} $$  How can I complete the proof from here?
107304
The problem is to show  $$\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),$$  for $t \gt 0$.  I'm pretty stuck. I thought about integration by parts and couldn't get anywhere with the integrand in its current form. I tried a substitution $u=e^{-x}$ and came to a new integral (hopefully after no mistakes)  $$ \int_0^1 \frac{u^{t-1}-1}{\log(u)}du, $$  but this doesn't seem to help either. I hope I could have a hint in the right direction... I really want to solve most of it by myself.  Thanks a lot!