2 Weird questions

Question

Seems like I'm full of weird mathematical questions!

Last time I made a question about imaginary numbers.

This time I have 2 seemingly unrelated questions. But nevertheless it's always good (and fun) to just ask away!

1) We know that, $f(x)=x^x$ (with $x$ being a Real number) is a non-integrable function. Therefore $∫x^x \, \mathrm dx$, has no algebraic solutions. The solutions exists, but they are just not algebraic. My question is: Ok, and if these solutions are not algebraic then what are they? Are they part of what we call transcendental functions? And if not what are they called and how do they look like?

2) We expanded algebra by axiomatically accepting $i^2$ to be equal to $-1$. Well, what if we do the same with $\ln(-1)$? Has that ever been done by now? And if yes can we use it in Euler's identity like so:

let's call $\ln(-1)$ as "$λ$" therefore $λ=\ln(-1)$

so from Euler we have

$e^{iπ}=-1 \Longleftrightarrow$

$\ln(e^{iπ})=\ln(-1) \Longleftrightarrow$ ?

$λ=iπ \Longleftrightarrow$

$π=λ/i$

?

And if yes, then what would that even mean? (btw if you google "$\ln(-1)$" the google calculator gives the following solution "$3.14159265\ldots\times i$" ?)

Answer 1

$1)$ Here's a solution to the first question $$\int_1^x t^t dt$$

This is an anti-derivative of the function $f(x)=x^x$.

$2)$ $\ln(-1)$ is already extended by anjoining $i$, or at least in a sense. Let $z=re^{i\theta}$ for $0 \leq \theta <2\pi$. Then we can define a complex logarithm as $\text{Log}(z)=\ln(r) + i\theta$. Then $\text{Log}(-1)=\text{Log}(1\cdot e^{i\pi})=\ln(1)+i\pi=i\pi$. Note my use of "a complex logarithm", because more than one can be defined. Since $z=re^{i\theta}=re^{i(\theta+2\pi)}$, it's clear how this would be done.

Answer 2

$1)$ WolframAlpha did not give me the full solution, but you can get it to give you the first 24 terms by pressing 'more digits'.

$$\int x^xdx=x+\frac{2\log(x)-1}4x^2+\frac{9\log^2(x)-6\log(x)+2}{54}x^3+O(x^4)$$

$2)$ Start with $(-1)^{-1}=(-1)^1=-1$. Thus, it is sufficient enough to show $\ln(-1^1)=\ln(-1^{-1})$.

In other words, $\ln(-1)=-\ln(-1)$. Since you have found $\ln(-1)=\frac\pi i=-\pi i$, and google proceeds to say $\ln(-1)=\pi i$, both are actually correct since $\ln(-1)=-\ln(-1)$

Answer 3

Suppose that we saw that

$$\frac{d}{dx}x^x=x^x(\log(x)+1)$$

$$\frac{d^2}{dx^2}x^x=x^x\left[(\log(x)+1)^2+\frac1x\right]$$

And we concluded the $n$th derivative of $x^x$ involved some $x^x$. Suppose we thought the same was true for the antiderivative:

$$\int_1^xt^tdt=f(x)x^x$$

Differentiating both sides with respect to $x$, we get

$$x^x=x^x\left[f'(x)+f(x)(\ln(x)+1)\right]$$

Dividing out the $x^x$, we get a differential equation

$$1=f'(x)+f(x)(\ln(x)+1)$$

Thus, if $f(x)$ were algebraic, $\ln(x)$ would be algebraic, leaving $f(x)$ to be transcendental.

However, this form is good for taking expansions and approximations, if that is of any interest. For example

$$\int_1^1t^tdt=f(1)1^1\implies f(1)=0$$

$$1=f'(1)+f(1)(\ln(1)+1)\implies f'(1)=1$$

Differentiat our DE and substitute $x=1$ to see that

$$0=f''(1)+f'(1)(\ln(1)+1)+\frac{f(1)}1\implies f''(1)=-1$$

So, by Taylor's theorem, we have

$$f(x)\approx(x-1)-\frac12(x-1)^2-\frac16(x-1)^3$$

One could also use Riemann sums.

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One can also see that if $\displaystyle\int_1^xt^tdt$ were algebraic, its derivative would be too, but it is not.