Three other transcendental(?) numbers

Question

I have $3$ numbers which seem to be transcendental, but probably very difficult to be proven transcendental. PARI/GP gives no indication that one of the numbers is algebraic.

The first one

$$\int_0^1 \ln(x^2+1) dx=\ln(2)+\frac{\pi}{2}-2$$

This number has the same numerical status as $\ln(2)+\frac{\pi}{2}$, which is the sum of two transcendental numbers. Of course, this does not even rule out, that it is rational.

The second one is the number $u$ with $$\int_0^u \ln(x^2+1)dx=1$$

Its numerical value is $1.709218728797693846135993\cdots$

It is the solution of the equation $$x\ln(x^2+1)-2x+2\arctan(x)=1$$

The third number is the number $v$ such that $$\int_0^v \ln(x)\tan(x) dx=0$$

Its numerical value is $1.414248858855349243230845167\cdots $. It is curious that the first five digits (including the $1$ belonging to the integer part) coincide with $\sqrt{2}$. This will be the hardest case because $\ln(x)\tan(x)$ seems to have no closed-form antiderivate.

Any ideas ?

Answer 1

We can prove that the first two numbers are transcendental. Here, we use Baker's theorem:

[Baker's Theorem]

If $\alpha_1, \ldots, \alpha_n$ are algebraic numbers, not $0$ or $1$. If $\log\alpha_1, \ldots, \log\alpha_n$ are linearly independent over $\mathbb{Q}$, then $1, \log\alpha_1, \ldots, \log\alpha_n$ are linearly independent over $\overline{\mathbb{Q}}$.

1. The first number: $\log 2 + \frac {\pi}2 - 2$.

   It is enough to show that $\log2 + \frac{\pi}2$ is transcendental. We use $\log(-1)=\pi i$. Thus, $\log(-1)$ and $\log 2$ are linearly independent over $\mathbb{Q}$. Hence, $1, \log(-1), \log2$ are linearly independent over $\overline{\mathbb{Q}}$. This shows that $\log2+\frac{\pi}2$ cannot be algebraic, hence transcendental.

2. The second number: A root of $x\log(x^2+1)-2x+2\arctan(x)=1$.

   Let $u$ be the positive real root of the equation $x\log(x^2+1)-2x+2\arctan(x)=1$. Suppose on the contrary that $u$ is algebraic. Then we have a $\overline{\mathbb{Q}}$-linear combination of $1$ and the logarithms of algebraic numbers resulting in zero: $$ u \log(1+u^2) -2u-1+ 2\frac1i \log \frac{1+ui}{\sqrt{1+u^2}}=0. $$ The last expression is due to $\tan\theta=u$ iff $e^{i\theta} = \frac{1+ui}{\sqrt{1+u^2}}$. By Baker's theorem, this would imply that $\log(1+u^2)$ and $\log \frac{1+ui}{\sqrt{1+u^2}}$ are linearly dependent over $\mathbb{Q}$. This is impossible since the former is real and the latter is imaginary. Therefore, the number $u$ must be transcendental.