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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 "$3.14159265\ldots\times i$" ?)

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    $\begingroup$ It's probably better to ask unrelated questions as separate questions, and use more descriptive question titles. $\endgroup$
    – littleO
    Jul 17 '16 at 2:42
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    $\begingroup$ For the second question, read about the complex logarithm where in fact $\log(-1)$ (and also $\text{Log}(-1)$) is defined and treated. $\endgroup$
    – JMoravitz
    Jul 17 '16 at 2:53
  • $\begingroup$ Given that "transcendental" and "not algebraic" mean the same thing, it seems confusing to me that you ask whether a function you know not to be algebraic is transcendental. What is your intention in asking such a question? (In terms that tend to be more familiar, it's kind of like asking whether a number that you know not to be rational is irrational) $\endgroup$ Jul 17 '16 at 3:52
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    $\begingroup$ Is it just me or someone just disliked my question? Are you guys for real? What's the point of having a page like this if people don't like questions? -.- No matter how dumb a question might appear to be, there is always something that someone can gain from it... I don't get it... Some people should get a life... $\endgroup$
    – Nuke
    Jul 18 '16 at 1:17
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    $\begingroup$ I didn't downvote, but the downvote was most likely because of the issues I mentioned, not because anyone thought these were dumb questions. $\endgroup$
    – littleO
    Jul 18 '16 at 2:41
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$1)$ 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.

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  • $\begingroup$ Hello Jason thanks for the quick answer! So... $\int$ $x^x$ $dx$ = $\int_1^x$ $t^t$ dt ? Isn't $\int_1^x$ $t^t$ $dt$ a definite integral as opposed to $\int$ $x^x$ $dx$ which is an indefinite integral? I'm not saying you are wrong. But I just don't understand it, that's why I ask. $\endgroup$
    – Nuke
    Jul 18 '16 at 2:29
  • $\begingroup$ ... When you showed that Log(-1) = iπ you used Log with base 10? Can I do the same with base e? Or that Log(-1) means that I can do the same with every base? So that means that Ln(-1)=iπ? Also, does that have any special meaning for our extended number system? I'm asking here, because I found nothing by googling these questions! $\endgroup$
    – Nuke
    Jul 18 '16 at 2:29
  • $\begingroup$ Also, if Log(-1)=iπ that means that i=Log(-1)/π and π=Log(-1)/i ? Maybe not huh? $\endgroup$
    – Nuke
    Jul 18 '16 at 2:31
  • $\begingroup$ @Nuke $\int_1^x t^t dt$ is a definite integral, but because its bounds depend on $x$, it is a function of $x$. By the First fundamental theorem of calculus, this is guaranteed to be an antiderivative of $f(x)=x^x$. In a sense, one can define $\int x^x dx$ as $\int_a^x t^t dt$ for any real $a$ in the domain of $f$. $\text{Log}$ is not $\log_{10}$, hence my use of a capital "L". It's a new function that I defined in a specific way. This new function agrees with $\ln$ for any positive real number. $\endgroup$
    – JasonM
    Jul 18 '16 at 2:35
  • $\begingroup$ @Nuke The fact that $\text{Log}(-1)=i\pi$ certainly has some meaning in physics or other applications. Mostly, $\text{Log}$ is just a complex-valued function that can be used to invert a complex exponential, but with the $[0, 2\pi)$ constraint on the imaginary part. $\endgroup$
    – JasonM
    Jul 18 '16 at 16:23
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$1)$ 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 $\ln(-1)=-\ln(-1)$

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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.


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

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