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1

The natural definition of $e$ is not a definition of that single number, but rather of a special function $x\mapsto \exp(x)$, which has the property that $\exp(0)=1$ and that it is its own derivative: $\exp'(x)=\exp(x)$. It turns out, that $\exp$ is uniquely determined by this. It follows that for any constant $k$, $\exp'(kx)=k\exp(kx)$ and ...


3

The most intuitive explanation I know involves a combination of three facts: If a particle has position $p(t)$ proportional to its velocity $p'(t)$, say $p'(t) = kp(t)$, then $p(t) = A e^{kt}$ for some constant $A$. We can take this as a definition of the exponential. In the complex plane, multiplication by $i$ is the same as a counter-clockwise rotation ...


8

Just think about it this way: $\pi$ is related to the circle, whose equation is $x^2+y^2=r^2$. Euler's constant e is related to the hyperbola, whose equation is $x^2-y^2=r^2$. In order to turn $y^2$ into $-y^2$ we need a substitution of the form $y\mapsto iy$.


1

For a given $y>0$, the equation $x\ln(y) = y\ln(x)$ is equivalent to $\ln(x)/x=\ln(y)/y$. Here is the graph of the function $f(x)=\displaystyle\frac{\ln(x)}{x}$. The function has a global maximum at $x=e$. If $y\leq 1$, then $f(y)\leq0$ and the equation has only one solution: $x=y$. If $y>1$, then $f(y)>0$ and $y\neq e$ and the equation has ...


1

Here is a relevant paper. They find a parametric equation for the non-$“x=y”$ branch of the graph (Desmos link), and then prove that the branches intersect at $(e,e)$. The rest of this answer is my own input. The fact that the two branches of the graph of $x^y=y^x$ intersect at $(e,e)$ is related to the fact that the graphs of $y=x^e$ and $y=e^x$ are ...


0

This is a derivation of Claude's answer. The definition of the Lambert-$W$ function is the solution to: $$x = y e^y$$ Our equation is: $$x \ln y = y \ln x$$ $$ \implies - \frac {\ln x}{x} = -\frac {\ln y}y = - \ln y e^{- \ln y}$$ $$\implies W\left({-\frac {\ln x}{x}}\right) = - \ln y = - y \frac{\ln x}{x}$$ $$\implies y = - \frac{x}{\ln ...


2

For most points on $y=x$, you have $dy/dx=1$. There is one point where $dy/dx$ could be $1$ or $-1$. Call that point $(x,y)=(a,a)$. $$x\ln y=y\ln x\\ \ln y+\frac xy\frac{dy}{dx}=\frac{dy}{dx}\ln x+\frac yx\\ \ln a+\frac{dy}{dx}=\frac{dy}{dx}\ln a+1$$ This has a unique solution unless $\ln a=1$


2

Any equation which can be written $$A+Bx+C\log(D+Ex)=0$$ has solutions expressed in terms of Lambert function. In the case of $x^y=y^x$, this writes $$y = -\frac{x}{\log x}\,W\left(-\frac{\log x}{x}\right)$$ and what Lambert and Euler showed is that, in the real domain, the function $W(z)$ exists if $z\geq -\frac 1e$. So, for the argument $-\frac{\log ...


8

Here is an approach. We may observe that we have $$ \int_0^{+\infty}\frac 1{(x+t)(\log^2 x +\pi^2)}dx=\frac 1{\log t}+\frac 1{1-t},\quad 0<t<1.\tag1 $$ Proof. Let $t$ be any real number such that $0<t<1$. Set $$ f(z):=\frac 1{(z+t)(\log z +i\pi)} $$ where $\log$ is the determination of the logarithm defined on the complex plane cut along ...


6

From the standard integral representation of the Euler-Mascheroni constant we have: $$ \gamma = \int_{0}^{1}\left(\frac{1}{\log(1-x)}+\frac{1}{x}\right)\,dx \tag{1}$$ hence if we define the Gregory coefficients $C_n$ through the Taylor series of $\frac{z}{\log(1-z)}$ in a neighbourhood of zero: $$ \frac{z}{\log(1-z)}=\sum_{n\geq 0}C_n\,x^n \tag{2}$$ we have ...



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