You are approximating the circle with the sum of the areas of triangles that fit inside. This is actually the first numerical way pi was calculated, done by Archimedes about 2300 years ago. Yes you can make the triangular slices thinner and thinner, in a process where you can fit in triangles without bound, however because they are getting thinner in proportion with the rate at which you are making more of them, the sum of the areas of the triangles actually stays finite.
 You do not fill the circle with copies of its radii (which will only have linear length and not width and so will not have any area). You fill the circle with tiny triangle slivers whose altitudes are nearly the same of the height and whose bases are really really small.
And you don't fill the circle with infinitely many. You fill the triangle with a lot of them but always so finite number even though it could be very large. And you reason what the area becomes when you take more and more of them.
And we don't fill the circle with them we almost fill the circle with them and we estimate.
SO for example if we fill in with $n$, a huge number, of triangle slivers, the height of on of this slivers will be nearly equal to the radius $r$ (just a teeny bit less) and base of these triangles, since there are $n$ of them and they are fitted along the circumference of the circle will have a $b$ very nearly equal to the the circumference divided by the number of triangles, that is $\frac Cn$.
So a triangle wedge has base = $b= \frac Cn$ and has a height $h = r$. So the area of the wedge is $\frac 12\cdot \frac Cn \cdot r$. And there are $n$ of them so the total area is $\require{cancel} n\cdot \frac 12 \cdot \frac Cn \cdot r = \cancel n\cdot \frac 12 \cdot \frac C{\cancel n}\cdot r = \frac 12 C\cdot r$. (Approximately. The larger $n$ is the closer this will be to the value and in a circle it will be reached.)
And $\frac {Area}{Radius} = \frac {\frac 12 Cr}{r} = \frac 12 C$.
Which considering $C = \pi \cdot diameter = 2\pi r$ is exactly what we would hope for.
Likewise the formula for the area is what we are used to if we consider $Area = \frac 12 C \cdot r = \frac 12 (\pi \cdot diameter)\cdot r= \frac 12(\pi \cdot 2r)\cdot r = \pi r^2$
 You could just as easily claim that, if a square of side length $L$ is positioned with its sides horizontal and vertical, infinitely many line segments parallel to the leftmost side can fit in the square, and therefore its area is infinite rather than $L^2$. The problem in both arguments is line segments have zero area, and infinitely many zeroes don't have an infinite sum. Having said that, we also, of course, can't say the area is $0$ either. So what's going on here?
The reason a line segment has zero area is because every positive number is an upper bound on its area, because it can be made part of an arbitrarily thin rectangle. If the square is divided into $n$ strips of width $L/n$, each has area $L^2/n$, and the total is $L^2$. Admittedly, "using a rectangle's area to compute a square's area" may sound circular; but the area axioms include a rectangle's area, as well as the "subset has at most a superset's area" principle I used to bound a line segment's area.
Now let's divide a circle into $n$ sectors with $n$ equally spaced radii, so each sector subtends $2\pi/n$ radians. Each sector's vertices are those of a triangle that's a subset of the sector, so the sector has at least the triangle's area, $\tfrac12 r^2\sin(2\pi/n)$. So for all positive integers $n$, the circle's area is at least $\tfrac12nr^2\sin(2\pi/n)$. But this doesn't grow in proportion to $n$, the way an "it contains infinitely many radii so has infinite area" intuition may expect. This approaches $\pi r^2$ as $n\to\infty$. Or you can use the OP's diagram's "rearrange them into an approximate rectangle argument, thereby avoiding the above trigonometric limit.

Since we can fill a circle with infinitely-many copies of its radius, isn't $\frac{\text{Circle area}} {\text{Circle radius}} =\infty?$

For three events to be independent, knowledge of the others should not affect the probability of each event. For pairwise independence, knowing one other event tells you nothing. But knowing both other events might shift the odds.
For example: draw 3 overlapping circles. Put a 1 in the 3 way overlap. Put 7 in each of the 2-way overlaps that are not in the center. Put a 1 in each of the regions belonging to exactly one set. Put a 7 in the complement of the union. The total is 32; each set contains 16, each 2-way overlap contains 8, so for example $A$ and $B$ are independent since $P(A) = \frac{1}{2}, P(B) = \frac{1}{2},$ and $P(A\cap B) = \frac{1}{4}$. Likewise for the other pairs, so we have full two way independence. But 3-way independence fails: $\frac{1}{8} = P(A)P(B)P(C) ≠ P(A\cap B \cap C) = \frac{1}{32}$.
Now what does this mean? Suppose I know $A$ is true. $P(C)$ is unaffected, since half of the elements of $A$ are in $C$.
Now uppose I know $A$ is true and $B$ is true. This makes $P(C)$ drop from $\frac{1}{2}$ to $\frac{1}{32}$. Knowing one of the other truth values told you nothing, but knowing both of them does give you information that changes the probability.

Intuitive Understanding of Independence of 3 Events

Typically associativity of the integral is proved early on. If $X$ is a semimartingale and the integral $K \cdot X = \int K dX$ makes sense then the integral $(HK) \cdot X = \int HK dX$ makes sense if and only if the integral $H \cdot (K\cdot X) = \int H d(\int K dX)$ makes sense, in which case they are equal. In "differential form", this is $H\, d(K dX) = (HK) dX$. I am assuming you are already comfortable with linearity of stochastic integrals.
Thus taking $dN_t = N_t r dt + N_t\alpha dB_t$ and integrating $1/N_t$ with respect to the semimartingales defined by either side gives
$$
(1/N_t) dN_t = (1/N_t) d(N_t r dt + N_t\alpha dB_t) = (1/N_t) d(N_t r dt)+ (1/N_t)d( N_t\alpha dB_t) = r dt + \alpha dB_t.
$$
There is no need to use Ito here, it is simply associativity of the integral. The key idea is that we are not "dividing by $N_t$", instead we are integrating $1/N_t$ with respect to two (equal) semimartingales. In differential form, it just looks like dividing.
 I find notation such as $\frac{dN_t}{N_t}=rdt+\alpha dB_t$ incredibly unfortunate, it leads to the type of confusion you have encountered (and that I used to encounter myself).
That is why I advise against using short-hand notation (at least until one becomes very confident in stochastic calculus techniques).
In long-hand notation:
$$N_t=N_0+\int_{h=0}^{h=t}rN_hdh+\int_{h=0}^{h=t}\alpha N_hdB_h$$
From the above, it should be obvious that $N_h$ cannot be taken out of the integral and brought to the LHS (which is not obvious in the short-hand notation). Ito's Lemma needs to be used, applied to $ln(N_t)$ as you point out. Let $F(N_t,t):=ln(N_t)$, then taking the derivatives:
$$\frac{\partial F}{\partial N_t}=\frac{1}{N_t}, \frac{\partial^2 F}{\partial N_t^2}=\frac{-1}{N_t^2}, \frac{\partial F}{\partial t}=0$$
Then:
$$F_t=F(N_0)+\int_{h=0}^{h=t}\left(\frac{\partial F}{\partial h}+\frac{\partial F}{\partial N_h}rN_h+0.5\frac{\partial^2 F}{\partial N_h^2}\alpha^2 N_h^2\right)dh+\int_{h=0}^{h=t}\frac{\partial F}{\partial N_t}\alpha N_hdB_h=\\=ln(N_0)+\int_{h=0}^{h=t}\left(r-0.5\alpha^2 \right)dh+\int_{h=0}^{h=t}\alpha dB_h=\\=ln(N_0)+(r-0.5\alpha^2)t+\alpha B_t$$
Because $F_t=ln(N_t)$, the final answer is (by exponentiating both sides):
$$N_t=N_0e^{(r-0.5\alpha^2)t+\alpha B_t}$$

A question about SDE and geometric Brownian motion.

<blockquote>
I have seen natural deduction proofs that are very complicated, or are
written in a way such that I suspect that there is some motivation
that I am surely missing. But I can't figure out how to get the
intuition/motivation for natural deduction proofs in general.
</blockquote>
The following proof is based on a form of natural deduction. Simpler than most, it may help. It uses direct proof (lines 8 and 9) and proof by contradiction (line 6).
Note: '=&gt;' is left associative here, so I am proving the equivalent of [~A =&gt; ~B] =&gt; [B =&gt; A].
Screen shot from my proof checker:
<img src="https://i.stack.imgur.com/9obuf.png" alt="enter image description here">

$\vdash (\lnot \chi \to \lnot \theta) \to (\theta \to \chi) $

Let $\mathbf {x, y} \in \mathbb{R}^n$ be unit vectors. 
Then, the angle $\theta_1$ between them is given by $ \cos \theta_1 = \mathbf{x^\top y}$.
Also, angle between $\mathbf{Ax}$ and $\mathbf{Ay}$ is given by :
$\cos \theta_2 = \dfrac{(\mathbf{Ax})^\top (\mathbf{Ay})}{\|\mathbf{Ax}\|\|\mathbf{Ay}\|}$. 
Now, $\|\mathbf{Ax}\|_2^2 = (\mathbf{Ax})^\top(\mathbf{Ax}) = \mathbf{x^\top A^\top Ax} = \mathbf{\lambda ^2 x^\top x}$. 
Thus, $\mathbf{\|Ax\|} = \mathbf{|\lambda|}$. 
$\implies \cos \theta_2 = \mathbf{\dfrac{\lambda^2 x^ \top y}{\lambda^2}} = \mathbf{x^ \top y} \implies \theta _1 = \theta _2$. 
Thus, $\mathbf{A}$ preserves angles and is a similarity transformation.

Why does an affine transformation $A$ when constrained by $A^TA=\lambda^2I$ result in a similarity transformation?

He explains his reasoning with respect to the first inequality in the paragraph above the inequality. Since each integer can be uniquely expressed as the product of a square-free integer and a square, each term on the right-hand sum occurs as a product of a single term from the square-free sum and a single term from the square sum. The inequality isn't strict because as $k$ and $j$ approach $n$, their product may exceed $n$, yielding some additional terms in the sum.
Similarly, in the second inequality, if you simply expand the product on the left, you will get all of the terms in the right-hand sum, and possibly some additional terms that you can throw away.
 Each $m&lt;n$ can be uniquely expressed as $m=kj^2$ where $k$ is a square-free integer, therefore
$$ \sum_{m&lt;n}\frac{1}{m}=\sum_{\substack{k,j \\ kj&lt;n}}\frac{1}{kj^2}\leqslant\sum_{k,j&lt;n}\frac{1}{kj^2}=\left(\sum_{k&lt;n}'\frac{1}{k}\right)\left(\sum_{j&lt;n}\frac{1}{j^2}\right) $$
where $k$ is a square free integer in the second and the third sum. As for the last inequality, let $p_1,\ldots,p_r$ be the prime numbers $&lt;n$, then developping the product gives
$$ \prod_{p&lt;n}\left(1+\frac{1}{p}\right)=\sum_{I\subset[\![1,r]\!]}\prod_{i\in I}\frac{1}{p_i}$$
Each square-free $k&lt;n$ has their prime divisors among $p_1,\ldots,p_r$ and thus is of the form $\prod_{i\in I}p_i$ where $I\subset[\![1,r]\!]$, therefore
$$\prod_{p&lt;n}\left(1+\frac{1}{p}\right)\geqslant\sum_{k&lt;n}'\frac{1}{k} $$
 When you multiply out the product
$$\left(\sum_{k&lt;n}'\frac1k\right)\left(\sum_{j&lt;n}\frac1{j^2}\right)\,,\tag{1}$$
you get the sum of all possible terms of the form $\frac1{kj^2}$, where $k$ is a square-free integer less than $n$, and $j$ is any integer less than $n$. Every positive integer $m&lt;n$ can be expressed as the product of such a $k$ and $j^2$, so these product terms include every fraction $\frac1m$ for $m&lt;n$. In general, however, they will also include some fractions $\frac1m$ with $m\ge n$: if $n-1$ is square-free, for instance, we can take $k=j=n-1$ and get $\frac1{(n-1)^3}$. Thus, every term of $\sum_{m&lt;n}\frac1m$ appears in $(1)$ after the latter is multiplied out, and $(1)$ generally contains extra terms as well, so
$$\left(\sum_{k&lt;n}'\frac1k\right)\left(\sum_{j&lt;n}\frac1{j^2}\right)\ge\sum_{m&lt;n}\frac1m\,.$$
For the second inequality let $P=\{p_1,\ldots,p_\ell\}$ be the set of primes less than $n$. When the product
$$\prod_{p&lt;n}\left(1+\frac1p\right)=\prod_{i=1}^\ell\left(1+\frac1{p_i}\right)\tag{2}$$
is multiplied out, there is one term for each possible choice of one factor, $1$ or $\frac1{p_i}$, from each of the $\ell$ factors. Specifically, if $S\subseteq P$, one of the terms of the product is
$$\frac1{\prod_{i\in S}p_i}\,,\tag{3}$$
obtained by choosing $1$ when $i\notin S$ and $\frac1{p_i}$ when $i\in S$.
Every square-free integer less than $n$ is a product of distinct primes less than $n$, so every term of $\sum\limits_{k&lt;n}'\frac1k$ is one of the terms $(3)$. However, there may also be some $S\subseteq P$ such that $\prod_{i\in S}p_i\ge n$, so the terms of $(2)$ after it’s multiplied out include all of the terms of $\sum\limits_{k&lt;n}'\frac1k$ and possible some others as well. Thus,
$$\prod_{p&lt;n}\left(1+\frac1p\right)\ge\sum_{k&lt;n}'\frac1k\,.$$

unexplained inequalities from Niven's 1971 proof $\sum 1/p$ diverges

The close-form result can be expressed as
$$\color{blue}{ \int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx = -nG+\frac\pi2 n \ln 2 
+ \pi \sum_{k=1}^{[\frac n2]}\ln \cos\frac{(n+1-2k)\pi}{4n} }
$$
as shown below. Note that
\begin{align}
I_n =
\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx
\overset{x\to\frac1x} == \frac12 J_n - nG
\end{align} 
where $ \int_1^\infty\frac{\ln x}{1+x^2} \,dx=G$ and
$$J_n =\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx $$
Substitute
$$1+x^{2n} = \prod_{k=1}^{n}(1+e^{i\pi\frac{n+1-2k}n }x^2)
$$
and use the known result
$\int_0^\infty \frac{\ln(1+ax^2)}{1+x^2}dx= \pi\ln(1+a^{\frac12})
$ to integrate
\begin{align}
J_{n}&amp; =\int_0^\infty\frac{dx}{1+x^2} \sum_{k=1}^{n}
\ln (1+e^{i\pi\frac{n+1-2k}n }x^2)
= \pi\sum_{k=1}^{n} \ln (1+e^{i\pi\frac{n+1-2k}{2n} })\\
 &amp;=n \pi \ln 2 + 2\pi\sum_{k=1}^{[\frac n2]} \ln \cos\frac{(n+1-2k)\pi}{4n} 
\end{align}
where the symmetry of the sequence is recognized in the last step.
 Not a complete answer, but an elaboration of what I said in the comments.
To evaluate $$\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx$$
Write it as $$\int_1^\infty\frac{\ln\left(x^{-2n}+1\right)+\ln\left(x^{2n}\right)}{1+x^2} dx = \int_1^\infty\frac{\ln\left(x^{-2n}+1\right)}{1+x^2} dx+2n\int_1^\infty\frac{\ln\left(x\right)}{1+x^2} dx$$
The second integral is a standard integral for Catalan's constant. To solve the first, make the substitution $x \to \frac{1}{x}$ to get $$\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx + 2nG$$
So if you know the value of any of the integrals with bounds $(0,\infty)$ or $(1,\infty)$ or $(0,1)$, you could find the other two.
<hr>
The series for $\ln(1+x)$ converges for $|x|&lt;1$, so expand it to get $$\int_0^1\frac{\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}x^{2nk}}{1+x^2}dx = \sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}\frac{x^{2nk}}{1+x^{2}}dx$$
Then you can expand $\frac{1}{1+x^2}$ to get $$\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\int_{0}^{1}x^{2nk}\sum_{m=0}^{\infty}\left(-1\right)^{m}x^{2m}dx = \sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\sum_{m=0}^{\infty}(-1)^m\int_{0}^{1}x^{2nk}x^{2m}dx$$
Finally, evaluate the inner integral to get
$$\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}\sum_{m=0}^{\infty}(-1)^m\frac{1}{2nk+2m+1}$$
 Just to put in more closed form what @Varu Vejalla has evaluated
$I=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx$
$\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\frac{1}{2}\int_{-\infty}^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx+\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=I+\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx$
$\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx [t=\frac{1}{x}]=\int_0^1\frac{\ln(1+t^{2n})}{1+t^2} dt-\int_0^1\frac{\ln(t^{2n})}{1+t^2} dx=I+2nG$
$$I=\frac{1}{4}\int_{-\infty}^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx-nG$$
Branch points of $\log$ (roots of $1+x^{2n}$) are $e^{\frac{i\pi}{2n}},e^{-\frac{i\pi}{2n}}, e^{\frac{3i\pi}{2n}}, e^{\frac{-3i\pi}{2n}},... e^{\frac{(2n-1)i\pi}{2n}}, e^{-\frac{(2n-1)i\pi}{2n}} $
Next we integrate in the complex plane, closing the contour in the upper half-plane for the roots $e^{-\frac{(2k-1)i\pi}{2n}}$ and in the lower half-plane for the roots $e^{+\frac{(2k-1)i\pi}{2n}}$ - to integrate the single-valued function.
Finally we get
$$I=\frac{2\pi{i}}{4*2i}\log\Bigl((-i-e^{\frac{i\pi}{2n}})(i-e^{-\frac{i\pi}{2n}})...(-i-e^{\frac{(2n-1)i\pi}{2n}})((-i-e^{-\frac{(2n-1)i\pi}{2n}})\Bigr)-nG=$$
$$=\frac{\pi}{4}\log\Bigl((2+2\sin\frac{\pi}{2n})...(2+2\sin\frac{(2n-1)\pi}{2n})\Bigr)-nG$$
$$I=\frac{\pi}{4}\log\Bigl((1+\sin\frac{\pi}{2n})...(1+\sin\frac{(2n-1)\pi}{2n})\Bigr)+\frac{\pi{n}}{4}\log2-nG$$

Is there a known closed form solution to $\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx$?

Note
$$I=\int\limits_{0}^{1} \frac{\ln(1-x+x^2)}{\sqrt{x}(1+x)}{d}x
\overset{t=\sqrt x} = 2\int\limits_{0}^{1} \frac{\ln(1-t^2+t^4)}{1+t^2}dt
$$
Let
$$J(a) = \int_0^1 \frac{\ln(\frac14(1+t^2)^2 \sec^2a -t^2)}{1+t^2}dt
$$
with $J(0) = \int_0^1 \frac{2\ln\frac{1-t^2}2}{1+t^2}dt=
-2G$
$$J’(a)= \int_0^1 \left(\frac{\tan a}{t^2+2t\cos a+1} + \frac{\tan a}{t^2-2t\cos a+1} \right)dt=\frac\pi2 \sec a
$$
$$J(\frac\pi6) = \int\limits_{0}^{1} \frac{\ln(1-t^2+t^4)-\ln3}{1+t^2}dt=\frac12I-\frac\pi4\ln3
$$
Then
\begin{align}
I &amp;= \frac\pi2\ln3+2J(\frac\pi6)
= \frac\pi2\ln3+2J(0) +2\int_0^{\pi/6} J’(a)da\\
&amp;=\pi \ln3-4G
\end{align}
 $$I=\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}dx\overset{t=1/x}=\int\limits_{1}^{\infty} \frac{\log(1-t+t^2)}{\sqrt{t}(1+t)}dt-\int\limits_{1}^{\infty} \frac{\log(t^2)}{\sqrt{t}(1+t)}dt$$
$$I=\frac{1}{2}\bigl(\int\limits_{0}^{\infty} \frac{\log(1-t+t^2)}{\sqrt{t}(1+t)}dt-\int\limits_{1}^{\infty} \frac{\log(t^2)}{\sqrt{t}(1+t)}dt\bigr)\overset{x=\sqrt{t}}=$$$$=\frac{1}{2}\int\limits_{-\infty}^{\infty} \frac{\log(1-x^2+x^4)}{1+x^2}dx-4\int\limits_{1}^{\infty} \frac{\log(x)}{1+x^2}dx$$
Branch points of logarithm: $e^{\frac{\pi{i}}{6}}, e^{\frac{5\pi{i}}{6}}, e^{\frac{-\pi{i}}{6}}, e^{\frac{-5\pi{i}}{6}}$
For the first couple of roots we close the contour in the bottom half of the complex plane, for the second - in the upper-half.
$$I=\frac{2\pi{i}}{2*2i}\log\Bigl((-i-e^{\frac{\pi{i}}{6}})(-i-e^{\frac{5\pi{i}}{6}})(i-e^{\frac{-\pi{i}}{6}})(i-e^{\frac{-5\pi{i}}{6}})\Bigr)-4G$$
$$I=\frac{\pi}{2}\log\biggl(4(1+\sin{\frac{5\pi}{6}})(1+\sin{\frac{\pi}{6}})\biggr)-4G=\frac{\pi}{2}\log(3^2)-4G=\pi\log3-4G$$
 Svyatoslav showed that $$\int_{0}^{1} \frac{\log(1-x+x^{2})}{\sqrt{x}(1+x)} \, \mathrm dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx- 4 G.$$
The following is a slightly different way to show that $$\int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx = 2 \pi \log (3). $$
<hr>
First notice that $$ \begin{align} \int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx &amp;= \int_{-\infty}^{\infty} \frac{\log(1+\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx+ \int_{-\infty}^{\infty} \frac{\log(1-\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx \\ &amp;= \int_{-\infty}^{\infty} \frac{\log(1+\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx + \int_{\infty}^{-\infty} \frac{\log(1+\sqrt{3}u + u^{2})}{1+u^{2}} (- \mathrm du) \\ &amp;= 2 \int_{-\infty}^{\infty} \frac{\log(1+\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx \end{align}$$
In a previous answer I used contour integration to show that $$I(a,b,\theta) = \int_{-\infty}^{\infty} \frac{\log \left(a^{2}+2ax \cos \theta + x^{2}\right)}{x^{2}+b^{2}} \, \mathrm dx = \frac{\pi}{b} \, \log \left(a^{2}+2ab \sin \theta +b^{2} \right), $$ where $a, b &gt;0$ and $0 &lt; \theta &lt; \pi$.
Therefore, $$ \int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx = 2 I \left( 1,1, \frac{\pi}{6} \right) = 2 \pi \log(3).$$

Evaluate an absolute monster integral $\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$

Let $$g(z) = \sum_{n\in \mathbb{Z}} \frac{(-1)^n}{\sinh(z+n)}$$
$g$ has $2$ and $2\pi i$ as period, so an elliptic function. Poles of $g$, which are all simple, are exactly $0,1,\pi i,1+\pi i$ modulo periods, with residues $1,-1,-1,1$.
$\DeclareMathOperator{\sn}{sn}\DeclareMathOperator{\ns}{ns}$
Since $g$ has only simple poles, it is easier to express it as Jacobi elliptic function. Recall that $\ns(z,k)$ has periods $4K,2iK'$ and residues $1$ at $0$. Hence $g(z)$ must be of form
$$g(z) =a \ns(az,k) - a \ns(a(z-\pi i),k) + C$$
then it is easy to determine $K'/K = 2\pi, a = 2K, C=0$. This puts $g(z)$ in terms of Jacobi-elliptic function:
$$g(z) = 2K[\ns(2Kz, k) - k \sn(2Kz,k)] \qquad K' = 2\pi K$$
<hr>
Mathematica code to calculate $g(z)$ numerically:
<pre><code>g[z_] := Module[{m = InverseEllipticNomeQ[Exp[-2 Pi^2]], K}, 
 K = EllipticK[m]; 
 2 K (JacobiNS[2 K*z, m] - Sqrt[m]*JacobiSN[2 K*z, m])];
</code></pre>
This can give, within an eyeblink, value of any $g(z)$ to great precision. For example, try
<pre><code>N[g[1/3], 1000]
</code></pre>
 We can evaluate it in terms of Jacobi theta functions.
This identity is occasionally useful:
$$
\sum _{n=-\infty }^{\infty }{\frac {b{u}^{n}}{1-a{q}^{2\,n}}}=
\frac{b}{2i}\;
{\frac {
{\theta_1} \left( \frac{i}{2}\ln \left( au \right) ,q \right) {
\theta_2} \left( 0,q \right) {\theta_3} \left( 0,q
 \right) {\theta_4} \left( 0,q \right) }{{\theta_1}
 \left( \frac{i}{2}\ln \left( u \right) ,q \right) {\theta_1} \left( 
\frac{i}{2}\ln \left( a \right) ,q \right) }}
$$
provided $1 &gt; |u| &gt; |q^2|$ so that the series converges.
For this problem, write $\sinh$ in terms of $\exp$ to get the left side of this,
with
$$
a=e^{-2 z},\quad b=2e^{-z},\quad q=e^{-1},\quad u=-e^{-1}
$$
to get the answer
$$
\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}
=
{\frac {i{{\rm e}^{-z}}{\theta_2} \left( iz+i/2,{{\rm e}^{-1}}
 \right) {\theta_2} \left( 0,{{\rm e}^{-1}} \right) { 
\theta_3} \left( 0,{{\rm e}^{-1}} \right) {\theta_4}
 \left( 0,{{\rm e}^{-1}} \right) }{{\theta_2} \left( i/2,{
{\rm e}^{-1}} \right) {\theta_1} \left( iz,{{\rm e}^{-1}}
 \right) }}
$$
Note, despite all those $i$s in there, this is real when $z$ is real.

Closed form of $\sum_{n=-\infty}^\infty \frac{(-1)^n}{\sinh (z+n)}$?

From the inductive step we have $(1+x)^{l-1}=\frac1l\sum_{k=1}^lk\binom lkx^{k-1}$.
Thus$$\begin{align*}(l+1)(1+x)^l&amp;=(1+x)\left[(l+1)(1+x)^{l-1}\right]\\&amp;=(1+x)\left[\left(\frac{l+1}l\right)\sum_{k=1}^lk\binom lkx^{k-1}\right]\\&amp;=\left[\left(\frac{l+1}l\right)\sum_{k=1}^lk\binom lkx^{k-1}\right]+x\left[\left(\frac{l+1}l\right)\sum_{k=1}^lk\binom lkx^{k-1}\right]\end{align*}$$The coefficient of $x^{k},1\le k\le l-1$,$$a_k=\left(\frac{l+1}l\right)\left[(k+1)\binom l{k+1}+k\binom lk\right]=(k+1)\binom{l+1}{k+1}$$which matches with the statement we need to prove. Similarly you can check the coefficients of $x^{0},x^l$.

New posts