109
If you want to try to prove $\int\frac{\mathrm dx}x=\ln x + C $ (for $x \gt 0$), try the substitution
$$ \begin{align}
x &= e^u \\
\mathrm dx &= e^u \mathrm du
\end{align} $$
This substitution is justified because the exponential function is bijective from $\mathbb{R}$ to $(0,\infty)$ (hence for every $x$ there exists a $u$) and continuously ...
107
$\DeclareMathOperator\antilog{antilog}$
First, when Napier invented logarithms, his application was not the inverse of the exponential function. He was originally multiplying quantities called "sines", which are not what you're thinking when you see that word. These "sines" are vaguely similar to the positions of the arrow in Zeno's ...
105
I'll try to give a soft answer to what I see as the spirit of the question, which is not why you get exactly log, but why the behaviour is different when integrating $x^{k}$ for $k=-1$.
The way I see it is that the logarithm would actually be there for other powers of $x$ too, but it's "hidden" by the fact that in a geometric series one term "gobbles up" ...
100
$\sum\frac1{n!}$ is not that special.
$\lim_{n\to\infty}\left(1+\frac1n\right)^n$ is not really special.
$f'(x)=f(x)$ is a very simple differential equation, but unremarkable, really.
$\ln (x)$ is only marginally nicer than other logarithms, in that its derivative is $\frac1x$.
The fact that a single number connects all of these (and many, many others) ...
91
What about the Prime Number Theorem? The number of primes smaller than $x$ is denoted by $\pi (x)$ and you have
$$\pi (x) \sim \frac{x}{\log x}$$
85
Let's use your result :
$$\int x^a \, dx = \frac{x^{a+1}}{a+1} + C=\frac{x^{a+1}-1}{a+1} +C'$$
and take the limit as $\,a\to -1\,$ (since $\;x^{a+1}=e^{(a+1)\ln(x)}$) :
$$\lim_{a\to\,-1}\int x^a \, dx = C'+\lim_{a\to\,-1}\frac{e^{(a+1)\ln(x)}-1}{a+1}=C'+\ln(x)$$
Hoping this will help your intuition,
75
This works in any base because $\log_b (a^c) = c \log_b(a)$
The practical reason for using base $10$ was a little old fashioned: it allowed the use of tables of logarithms instead of a calculator and reducing these calculations to addition and subtraction. Calculators then provided a $\log_{10}$ function as a carry-over from tables and just called the ...
72
In mathematics, $\log n$ is most often taken to be the natural logarithm. The notation $\ln(x)$ not seen frequently past multivariable calculus, since the logarithm base $10$ finds relatively little use.
This Wikipedia page gives a classification of where each definition, that is base $2$, $e$ and $10$, are used:
$\log (x)$ refers to $\log_2 (x)$ in ...
69
$$\boxed{\displaystyle\int_0^1\log\log\left(\frac1x+\sqrt{\frac1{x^2}-1}\right)\mathrm dx=-\gamma-2\ln\frac{2\Gamma(3/4)}{\Gamma(1/4)}}\tag{$\heartsuit$}$$
Derivation:
After the change of variables $x=\frac{1}{\cosh u}$ the integral becomes
$$\int_0^{\infty}\ln u \frac{\sinh u}{\cosh^2 u}du,$$
as was noticed above by Eric. We would like to integrate by ...
answered May 13 '13 at 20:58
Start wearing purple
51.1k12 gold badges147 silver badges206 bronze badges
65
For $a > 0$, let $b = \frac12 + \frac1a$ and $I(a)$ be the integral
$$I(a) = \int_0^1 \frac{\log(a+x)}{\sqrt{x(1-x)(2-x)}}dx$$
Substitute $x$ by $\frac{1}{p+\frac12}$, it is easy to check we can rewrite $I(a)$ as
$$
I(a)
= -\sqrt{2}\int_\infty^{\frac12}\frac{\log\left[a (p + b)/(p + \frac12)\right]}{\sqrt{4p^3 - p}} dp
$$
Let $\wp(z), \zeta(z)$ and $\...
64
Let's take a closer look at your calculations:
$$x^6\geq x^8 \Leftrightarrow \ln(x^6)\geq \ln(x^8)$$
Here we must have $x\neq 0$, because $\ln(0)$ is not well-defined. So you have to check if the inequality holds for $x=0$ separately. (It does hold).
Now you want to use $$\ln(x^6)=6\cdot\ln(x)\geq 8\cdot\ln(x)=\ln(x^8).$$ This only holds for $x>0$, as ...
61
Observe that:
$$ \begin{align*}
\log_5 7 &= \dfrac{3}{3}\log_5 7 \\
&= \dfrac{1}{3}\log_5 7^3 \\
&= \dfrac{1}{3}\log_5 343 \\
&< \dfrac{1}{3}\log_5 625\\
&= \dfrac{1}{3}\log_5 5^4\\
&= \dfrac{1}{3}(4)\\
&= \sqrt{\dfrac{16}{9}}\\
&< \sqrt{\dfrac{18}{9}}\\
&= \sqrt{2}\\
\end{align*} $$
as desired.
58
Converting a comment to an answer (my third for this question!), by request. I think it might actually constitute my best suggestion.
Consider
$$b\stackrel{p}{\lrcorner}r$$
with $b$ the base, $p$ the exponent, and $r$ the result (for lack of a better word (see below)), with a fill-in-the-blank philosophy: whatever's missing is what the symbol represents.
$...
57
To evaluate $\log_8 128$, let
$$\log_8 128 = x$$
Then by definition of the logarithm,
$$8^x = 128$$
Since $8 = 2^3$ and $128 = 2^7$, we obtain
\begin{align*}
(2^3)^x & = 2^7\\
2^{3x} & = 2^7
\end{align*}
If two exponentials with the same base are equal, then their exponents must be equal. Hence,
\begin{align*}
3x & = 7\\
x & = \frac{7}{3}
\...
56
Aside from $r=1$, no. To prove it, suppose we had an example. Then we'd write $$\frac mn=e^{\frac ab}\implies e^a=\left( \frac mn \right)^b$$ But, with $a\neq 0$ this would tell us that $e$ was algebraic, which is not the case.
54
Take the upper bound:
$$
\ln {x} \leq x-1
$$
Apply it to $1/x$:
$$
\ln \frac{1}{x} \leq \frac{1}{x} - 1
$$
This is the same as
$$
\ln x \geq 1 - \frac{1}{x}.
$$
54
We want a function that changes multiplication into addition. That is, we want $$f(xy) = f(x) + f(y).\tag 1 $$
Substituting $y=1,$ we get $f(x) = f(x) + f(1),$ so we know that $f(1) = 0.$
Now, let's suppose that $f$ is differentiable. After all, we want to find as nice a function as possible. Let's hold $y$ constant for the moment, and differentiating $(...
53
Computing a Related Contour Integral:
Define
$$f(z)=\frac{i}{2}\frac{z-1}{1+az}\left(\frac{1}{z-\ln{z}+\ln\left(\frac{\pi}{2}\right)}+\frac{1}{z-\ln{z}+2\pi i+\ln\left(\frac{\pi}{2}\right)}\right)$$
and let $\gamma$ denote a keyhole contour deformed around $[0,\infty]$. Restricting the argument between $0$ and $2\pi$, it is not hard to see that $f(z)$ has ...
52
Note the only way $2^n+1$ can have one more digit than $2^n$ is if $2^n$ ended in a $9$ (actually ends is $\cdots 999999$ but that is not important). $2^n$ can never end in a $9$.
51
You are correct if we endow $\Bbb R_{> 0}$ with the strange vector space structure in which "addition" is given by the usual multiplication, and "scalar multiplication" is given by exponentiation. When people say that logarithms are not linear, they are usually thinking of giving $\Bbb R$ the usual vector space structure, and with this being understood, ...
50
First observe that
$$x^2-2 x+2 \log{(1+x)} = 2 \sum_{k=3}^{\infty} (-1)^{k+1} \frac{x^k}{k}$$
The integral is then equal to
$$2 \sum_{k=0}^{\infty} \frac{(-1)^k}{k+3} \int_0^1 dx \frac{x^k}{\sqrt{1-x^2}}$$
Now, we will need separate treatments for the even and odd terms (1):
$$\int_0^1 dx \frac{x^k}{\sqrt{1-x^2}} = \begin{cases} \frac{\displaystyle 1}{\...
50
By the substitution $x\mapsto x^{-1}$ it is not hard to see that
$$I(\nu^{-1},1)=\int^\infty_0\frac{\arctan^2{x}\arctan(\nu x)}{x^2}{\rm d}x$$
First, start off by re-expressing the integral
\begin{align}
\int^\pi_0x^2\cos(nx)\cot\left(\frac{x}{2}\right)\ {\rm d}x
&=-{\rm Re}\int_C\frac{z+1}{z-1}z^{n-1}\ln^2{z}\ {\rm d}z\\
&=(-1)^{n}\int^1_0\frac{1-x}...
49
This integral can be evaluated in a closed form for arbitrary real exponents, and does not seem to be related to Herglotz-like integrals.
Assume $a,b\in\mathbb{R}$. Note that
$$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\int_0^\infty\frac{\ln\left(\...
answered Mar 7 '14 at 2:12
Vladimir Reshetnikov
44.3k5 gold badges137 silver badges268 bronze badges
49
For a function to be a logarithm, it should satisfy the law of logarithms:
$\log ab = \log a + \log b$, for $a,b \gt 0$.
If it were the identity function, this would become $ab = a + b$, which clearly is not always true.
48
Here is another Feynman's way to evaluate the integral. Set $x=\frac1t$, then
$$
\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x}\ dx=\int_0^1\frac{t-1-t\ln t}{(1+t^2)\ln^2t}\ dt.
$$
Now consider
$$
\mathcal{I}(\alpha)=\int_0^1t^\alpha\cdot\frac{t-1-t\ln t}{(1+t^2)\ln^2t}\ dt.
$$
Hence
\begin{align}
\frac{d^2\mathcal{I}}{d\alpha^2}&=\int_0^1\...
47
No. $(\log_2(3))^2$ can't be simplified.
However, $\log_2(3^2)=2\log_2(3)$.
47
The formula $$\log_a b^c = c \log_a b$$ is true only if $b > 0$ (if we assume that $\log_a$ is a real-valued function). Therefore, an alternative method of solution can proceed as follows: $$\log (x+2)^2 = 2 \log 5 = \log 5^2 = \log 25,$$ and because now all the arguments to $\log$ on both sides must be positive, we have $$(x+2)^2 = 25$$ or $$(x+2-5)(x+...
46
Logarithms are all about solving exponential equations right? For example, logarithms are useful if you want to solve $2^x=3$. Maybe you are really confused about how in the world finding the area under $y=\frac{1}{x}$ has anything to do with solving such equations. Hopefully this post will illuminate that - but it may be a bit long winded.
Say you ...
45
First, we transform the integral into a more computable form by using some substitutions.
$$\begin{align*}\displaystyle Q &= \int_0^1\sqrt{\frac{2-x}{(1-x)\,x}}\,\log\left(\frac{(2-x)\,x}{1-x}\right)dx\\ &=\int_0^1 \sqrt{\frac{1+u}{u(1-u)}}\log \left( \frac{(1+u)(1-u)}{u}\right)du \quad \color{blue}{\text{where }u=1-x}\\
&= \int_0^1 \frac{1+u}{\...
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