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13

Define $I(a)=\int_0^{\infty} e^{-(x+1)a}\frac{\sin(x)}{x+1}dx$. Then $\displaystyle I'(a)=-\int_0^{\infty} e^{-(x+1)a}\sin(x)dx=-\frac{e^{-a}}{a^2+1}$, and since $\lim_{a\to \infty} I(a)=0$, we have $\ \displaystyle I(0)=\int_0^{\infty} \frac{\sin(x)}{x+1}=\int_0^{\infty}\frac{e^{-a}}{a^2+1}da=\int_0^1\frac{dx}{1+\ln^2 x}.$


12

Hint. You may observe that $$ \frac1{1+\ln^2 x}=-\Im \frac1{i-\ln x}=-\Im \int_0^{\infty}e^{-(i-\ln x)t}dt,\quad x \in (0,1), $$ giving $$ \begin{align} \int_0^1 \frac{1}{1+\ln^2 x}\,dx&=-\Im \int_0^1\!\!\int_0^{\infty}e^{-(i-\ln x)t}dt\:dx\\\\ &=-\Im \int_0^{\infty}\!\!\left(\int_0^1x^t dx\right)e^{-it}dt\\\\ &=-\Im ...


5

Differentiation under the integral sign gives: $$ \int_{0}^{1}\log(x)\log(1-x)\,dx = \left.\frac{\partial^2}{\partial\alpha\,\partial\beta}\int_{0}^{1}x^\alpha(1-x)^{\beta}\,dx\right|_{\alpha,\beta=0} $$ hence you just have to differentiate a beta function and $\zeta(2)$ arises as $\psi'(1)$. Differentiation is carried on through: $$ \frac{d}{dz}\,f(z) = ...


3

This has to do with the fact that using integration by parts you get a dilogarithm, and $\text{Li}_2(1)=\zeta(2)$, since the two defining series happen to be the same at $x=1$. Why $\zeta(2)$ involves $\pi^2$ can be seen as a by-product of the series expansion of $\sin(x)$, as can be read about here, on the solution to the Basel problem. In short one shows ...


3

Given $\log_{a}(x)\;,\log_{b}(x)$ and $\log_{c}(x)$ are in Arithmetic Progression. So Using if $a,b,c$ are in Arithemic Progression, Then $2b=a+c$ So $\displaystyle 2\log_{b}(x) = \log_{a}(x)+\log_{c}(x)\Rightarrow \frac{2}{\log_{x}(b)} = \frac{1}{\log_{a}(x)}+\frac{1}{\log_{x}(c)}$ So $\displaystyle \frac{2}{\log_{x}(b)} = ...


3

Then , $k(y\log x+x\log y)=k(xy^2+xyz-x^2y+x^2y+xyz-xy^2)=2kxyz$ Again , $k(z\log y+y\log z)=k(xyz+z^2y-zy^2+y^2z+xyz-z^2y)=2kxyz$. So, $k(y\log x+x\log y)=k(z\log y+y\log z)\implies x^yy^x=y^zz^y$ , as $k\not=0$. Similarly you can prove the second equality..


3

use $\log_a b=‎\dfrac{\log b}{\log a}$


3

Here's a way that may be the easiest to understand, using the change-of-base formula in its simplest form: $$ (\log_4 7)(\log_7 5) = \frac{\log_e 7}{\log_e 4} \cdot \frac{\log_e 5}{\log_e 7} = \frac{\log_e 5}{\log_e 4} = \log_4 5. $$ Here's a way that uses a corollary of the change-of-base formula: $$ \underbrace{(\log_4 7)(\log_7 5) = (\log_7 7)(\log_4 ...


2

If in doubt convert to powers so that $$7=4^a; 5=7^b=(4^a)^b=4^{ab}$$ and then extract $ab$ from this equation.


2

The simplest way to see this is by taking $$\lim_{x \to \infty} \frac{d}{dx}\ln x = \lim_{x \to \infty} 1/x = 0$$ and as such observing that because the slope approaches zero $\ln x$ flattens out as $x \to \infty$. Unfortunately, this method offers zero intuition. Similar behavior occurs in a discrete case with the harmonic series. A grouping technique ...


2

In compositional data analysis, the distance function $d(x,y) = |\ln(x) - \ln(y)| = \left| \ln\left( \frac{x}{y} \right) \right|$ is known as the log-ratio distance, and it was introduced by J. Aitchison. But also the log-normal distribution involves this distance function. If parameterized with the geometric mean $m$, the probability density function of the ...


2

The converse is actually false. For example, define $f(z)=\int_1^z\frac{dt}{t}+1$ in the open disc $D=\{z\in\mathbb{C}:|z-1|\lt 1\}$. It is clear that $\int_1^z\frac{dt}{t}$ is a branch of the logarithm on $D$, so $$ e^{f(z)}=e^{\int_1^z\frac{dt}{t}}e^1=ez\neq z. $$ However, the assertion is true with the additional assumption that $e^{f(t)}=t$ for some $t$ ...


2

Hint : You have the following equality $$1 = \frac{a}{b} \times \frac{a+b}{b}, $$ and observe that $\frac{a+b}{b} = \frac{a}{b}+\frac{b}{b} = \frac{a}{b} + 1$, now we define $Y=\frac{a}{b}$, to get that $Y$ is a solution of $$1 = Y\times(Y+1)\qquad \Leftrightarrow\qquad Y^2+Y-1=0. $$


1

A variant: $$ \int_1^{\infty}\frac{\sin(x-1)}{x}dx=\Im\int_1^{\infty}\frac{e^{i(x-1)}}{x}dx= \quad (1)\\ \Im\int_1^{\infty}\int_0^{\infty}e^{i(x-1)-xt}dtdx=\Im\int_0^{\infty}\int_1^{\infty}e^{i(x-1)-xt}dxdt=\quad (2) \\ \Im\int_0^{\infty}\frac{e^{-t}}{t-i}dt=-\int_0^{\infty}\frac{e^{-t}}{1+t^2}dt=\quad (3) \\ \int_0^1\frac{1}{1+\log^2(x)}dx\quad (4) $$ ...


1

Yes, it is true that $$a^{\log_g(b)}=b^{\log_g(a)}$$ You can see this by taking logs on both sides, and using $\log_g x^y=y \log_g x$. Or, if you want a direct proof, let $x:= \log_g a$ and $y=\log_g b$. This means $$a=g^x \, \mbox{ and } b=g^y$$ You want to prove that $a^y=b^x$. This is easy: $$a^y=(g^x)^y=g^{xy}=(g^{y})^x=b^x$$


1

It might be useful to present an approach to evaluating these limits that forgoes use of L'Hospital's Rule. Here, we begin by using the following definition of the exponential function $e^x$. $$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n \tag 1$$ Then, it is straightforward to show from $(1)$ that for $x>0$ $$e^x\ge \left(1+\frac x2\right)^2 \tag ...


1

Since $$\lim _{ x\rightarrow \infty }{ \frac { x }{ { b }^{ x } } =0,b>1\quad } $$ then for enough big $x$ : $$\frac { 1 }{ { b }^{ x } } <\frac { x }{ { b }^{ x } } <1$$ denote $b=e^{ \varepsilon }$ for small arbitrary $\varepsilon >0$ then we get: $$\frac { 1 }{ { e }^{ \varepsilon x } } <\frac { x }{ e^{ \varepsilon x } } <1$$ or ...


1

$$\lim\limits_{n\to\infty}\frac{(\log n)^{\frac12}}{\log n^{\frac12}}=2\lim\limits_{n\to\infty}\frac{\sqrt{\log n}}{\log n}$$ $$=2\lim\limits_{n\to\infty}\frac{1}{\sqrt{\log n}}=0$$


1

Although people have hinted at it, you must use the fact that any integer $n > 1$ has a unique prime factorization. In order for to obtain $9^a = 15^b$, we must have $3^{2a} = 3^b5^b$, or $3^{2a-b} = 5^b$. But by unique factorization, no positive power of $3$ can equal a positive power of $5$. Thus, $\log_9 15$ must be irrational. [Edited to account ...


1

HINT : We have $$\frac 1n\lt\frac mk\lt n\iff \frac mn\lt k\lt mn\tag 1$$ Setting $\lfloor\frac mn\rfloor=s$ gives $$(1)\iff s+1\le k\le mn-1$$with $s\le\frac mn\lt s+1$. Hence, one has $50=(mn-1)-(s+1)+1$.


1

Yes, if you are concerned with the function $(x,y) \mapsto \sin x \log |y-1|$. Let $y \neq 1$ and let $f: x \mapsto \sin x \log |y-1|$ on $\mathbb{R}$. Then $$ \int_{x} \sin x \log |y-1| = -\log|y-1|\int_{x} D\cos x = -\cos x \log|y-1| + \text{some constant}. $$ But if you are concerned with the function $x \mapsto (x, f(x)) =: (x,y) \mapsto \sin x \log ...



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