lebesgue integral - Condition for integrability on finite measure space

Let $(X,\mathcal{F},\mu)$ be a finite measure space. If $f:X\rightarrow \mathbb{R}$ is a measurable real function, show that, $f\in L^1(\mu)$ iff $\sum\limits_{n=0}^{\infty}\mu(\{f\geq n\})<\infty$. Am a bit stuck on the $(\Leftarrow)$ direction so any help is appreciated, thanks.

calculus - A sine integral $int_0^{infty} left(frac{sin x }{x }right)^n,mathrm{d}x$

The following question comes from Some integral with sine post

$$\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$$
but now I'd be curious to know how to deal with it by methods of complex analysis.
Some suggestions, hints? Thanks!!!

Sis.

Answer

Here's another approach.

We have
$$\begin{eqnarray*}
\int_0^\infty dx\, \left(\frac{\sin x}{x}\right)^n

&=& \lim_{\epsilon\to 0^+}
\frac{1}{2} \int_{-\infty}^\infty dx\,
\left(\frac{\sin x}{x-i\epsilon}\right)^n \\
&=& \lim_{\epsilon\to 0^+}
\frac{1}{2} \int_{-\infty}^\infty dx\,
\frac{1}{(x-i\epsilon)^n}
\left(\frac{e^{i x}-e^{-i x}}{2i}\right)^n \\
&=& \lim_{\epsilon\to 0^+}
\frac{1}{2} \frac{1}{(2i)^n} \int_{-\infty}^\infty dx\,
\frac{1}{(x-i\epsilon)^n}

\sum_{k=0}^n (-1)^k {n \choose k} e^{i x(n-2k)} \\
&=& \lim_{\epsilon\to 0^+}
\frac{1}{2} \frac{1}{(2i)^n}
\sum_{k=0}^n (-1)^k {n \choose k}
\int_{-\infty}^\infty dx\, \frac{e^{i x(n-2k)}}{(x-i\epsilon)^n}.
\end{eqnarray*}$$
If $n-2k \ge 0$ we close the contour in the upper half-plane and pick up the residue at $x=i\epsilon$.
Otherwise we close the contour in the lower half-plane and pick up no residues.
The upper limit of the sum is thus $\lfloor n/2\rfloor$.
Therefore, using the Cauchy differentiation formula, we find

$$\begin{eqnarray*}
\int_0^\infty dx\, \left(\frac{\sin x}{x}\right)^n
&=& \frac{1}{2} \frac{1}{(2i)^n}
\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k}
\frac{2\pi i}{(n-1)!}
\left.\frac{d^{n-1}}{d x^{n-1}} e^{i x(n-2k)}\right|_{x=0} \\
&=& \frac{1}{2} \frac{1}{(2i)^n}
\sum_{k=0}^{\lfloor n/2\rfloor}
(-1)^k {n \choose k}
\frac{2\pi i}{(n-1)!} (i(n-2k))^{n-1} \\

&=& \frac{\pi}{2^n (n-1)!}
\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}.
\end{eqnarray*}$$
The sum can be written in terms of the hypergeometric function but the result is not particularly enlightening.

combinatorics - How to determine the size of the complete game tree for basic [M]?

You can read the rules of the game here, or actually play it free on the mobile mbrane app, but it's not required to address the question.

Essentially: players take turns placing integers onto an empty Sudoku until no more integers may be legally placed.

Part of the complete gametree involves ~6.67x10²¹ complete Sudoku, reduced for rotation but not substitution because the integers have magnitude.

(Full disclosure: this part of the tree is almost entirely meaningless as strategic placement of the integers, influenced by the topology, seems to always result in incompletable Sudoku, which leads to the real problem.)

Here is an image to illustrate how the dead sectors occur, for those interested. (x's mark the dead sectors):

At some point I plan to figure out how to derive total number of broken Sudoku--dead sectors can be created with as few as 9 placements--but for now I just want to make sure I understand how the exponential expansion of placement sequence interacts mathematically with the factorial structure of Sudoku, and the proper notation.

• What is the size of the basic [M] gametree, assuming only completable Sudoku?

Alternately:

• How to derive the complete gametree size of basic [M] on a 2x2(2x2) gameboard?

The second example on 2x2(2x2) can actually be checked!

Sorry if this is a really basic question, but I only have basic maths, so be kind! (I'm working up to a question on how to determine the complexity class of the basic game:)

integration - Triple Euler sum result $sum_{kgeq 1}frac{H_k^{(2)}H_k }{k^2}=zeta(2)zeta(3)+zeta(5)$

In the following thread

I arrived at the following result

$$\sum_{k\geq 1}\frac{H_k^{(2)}H_k }{k^2}=\zeta(2)\zeta(3)+\zeta(5)$$

Defining

$$H_k^{(p)}=\sum_{n=1}^k \frac{1}{n^p},\,\,\, H_k^{(1)}\equiv H_k $$

But, it was after long evaluations and considering many variations of product of polylogarithm integrals.

I think there is an easier approach to get the solution, any ideas ?

Answer

Here's a derivation that, while fairly long, is self-contained and uses only basic series manipulation techniques, like partial fractions decomposition, telescoping, swapping the order of summation, etc. It leans heavily on ideas from Borwein and Girgensohn's paper "Evaluation of Triple Euler Sums" (Electronic Journal of Combinatorics 3(1) 1996).

First, some notation. Define the multiple zeta functions by

\begin{align}
\zeta_N(a) &= \sum_{x=1}^N \frac{1}{x^a}, \:\:\: \zeta_N(a,b) = \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^a y^b}, \:\:\: \zeta_N(a,b,c) = \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^b z^c}, \\
\zeta(a,b) &= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^a y^b}, \:\:\: \zeta(a,b,c) = \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^b z^c}.
\end{align}

We will need the following symmetry relation, as well as expressions for $\zeta(4,1)$ and $\zeta(2,2,1) + \zeta(2,1,2)$. Proofs for all of these are given at the end of the post.
\begin{align}
\zeta_N(a,b) + \zeta_N(b,a) &= \zeta_N(a) \zeta_N(b) - \zeta_N(a+b) \tag{1}\\
\zeta(4,1) &= \zeta(5) - \zeta(3,2) - \zeta(2,3) \tag{2}\\
\zeta(2,2,1) + \zeta(2,1,2) &= \zeta(2,3) + \zeta(3,2) \tag{3}

\end{align}

Given these, we have

The Main Proof:
\begin{align}
\sum_{k=1}^{\infty} \frac{H^{(2)}_k H_k}{k^2} &= \sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1} H_{k-1}}{k^2} + \sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1}}{k^3} + \sum_{k=1}^{\infty} \frac{H_{k-1}}{k^4} + \sum_{k=1}^{\infty} \frac{1}{k^5} \\
&= \sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1} H_{k-1}}{k^2} + \zeta(3,2) + \zeta(4,1) + \zeta(5).
\end{align}
The most complicated sum is the first, so let's look at that more closely.

\begin{align}
\sum_{k=1}^{\infty} \frac{H^{(2)}_{k-1} H_{k-1}}{k^2} &= \sum_{k=1}^{\infty} \frac{1}{k^2} \zeta_{k-1}(2) \zeta_{k-1}(1) \\
&= \sum_{k=1}^{\infty} \frac{1}{k^2} (\zeta_{k-1}(2,1) + \zeta_{k-1}(1,2) + \zeta_{k-1}(3)), \text{ by (1)} \\
&= \zeta(2,2,1) + \zeta(2,1,2) + \zeta(2,3), \text{ by definition of the multiple zeta functions} \\
&= 2\zeta(2,3) + \zeta(3,2), \text{ by (3)}.
\end{align}
Thus
\begin{align}
\sum_{k=1}^{\infty} \frac{H^{(2)}_k H_k}{k^2} &= 2 \zeta(2,3) + \zeta(3,2) + \zeta(3,2) + \zeta(5) - \zeta(3,2) - \zeta(2,3) + \zeta(5), \text{ by (2)} \\
&= \zeta(2,3) + \zeta(3,2) + 2 \zeta(5) \\

&= \zeta(2) \zeta(3) - \zeta(5) + 2 \zeta(5), \text{ by (1)} \\
&= \zeta(2) \zeta(3) + \zeta(5).
\end{align}

---

---

Proof of (1):
\begin{align}
\zeta_N(a,b) + \zeta_N(b,a) &= \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^a y^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a} \\
&= \sum_{y=1}^N \sum_{x=y+1}^N \frac{1}{x^a y^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a}, \\
& \:\:\:\:\:\: \text{ swapping the order of summation on the first sum} \\
&= \sum_{x=1}^N \sum_{y=x+1}^N \frac{1}{y^a x^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \frac{1}{x^b y^a}, \text{ relabeling variables on the first sum} \\

&= \sum_{x=1}^N \sum_{y=1}^N \frac{1}{y^a x^b} - \sum_{x=1}^N \frac{1}{x^{a+b}}, \text{ combining sums} \\
&= \zeta_N(a) \zeta_N(b) - \zeta_N(a+b). \square
\end{align}

Proof of (2):
\begin{align}
\zeta(4,1) &= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^4 y} \\
&= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{x^4 (x-y)}, \text{ reindexing the second sum} \\
&= \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \left(-\frac{1}{x^4 y} - \frac{1}{x^3 y^2} - \frac{1}{x^2y^3} - \frac{1}{x y^4} + \frac{1}{(x-y)y^4}\right), \\
&\:\:\:\:\: \text{ by partial fractions decomposition}\\

&= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \left(\frac{1}{(x-y)y^4} - \frac{1}{x y^4} \right) \\
&= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{x=1}^{\infty} \sum_{y=1}^{x-1} \frac{1}{y^4} \left(\frac{1}{x-y} - \frac{1}{x} \right) \\
&= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{y=1}^{\infty} \frac{1}{y^4} \sum_{x=y+1}^{\infty} \left(\frac{1}{x-y} - \frac{1}{x} \right), \\
& \:\:\:\:\: \text{ swapping the order of summation} \\
&= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \sum_{y=1}^{\infty} \frac{1}{y^4} \sum_{x=1}^y \frac{1}{x}, \text{ as the sum telescopes} \\
&= - \zeta(4,1) - \zeta(3,2) - \zeta(2,3) + \zeta(4,1) + \zeta(5) \\
&= \zeta(5) - \zeta(3,2) - \zeta(2,3). \square
\end{align}

For the proof of (3), we need the following additional symmetry result:

\begin{equation}
\zeta_N(a,b,c) + \zeta_N(a,c,b) + \zeta_N(c,a,b) = \zeta_N(c) \zeta_N(a,b) - \zeta_N(a,b+c) - \zeta_N(a+c,b) \tag{4}
\end{equation}

Proof of (4):
\begin{align}
&\zeta_N(a,b,c) + \zeta_N(a,c,b) + \zeta_N(c,a,b) \\
&=\sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^b z^c} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^a y^c z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b} \\
&= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=z+1}^{x-1} \frac{1}{x^a y^b z^c} + \sum_{y=1}^N \sum_{x=y+1}^N \sum_{z=1}^{y-1}\frac{1}{x^a y^c z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\
&\:\:\:\:\:\text{ swapping order of summation on the first two sums} \\

&= \sum_{z=1}^N \sum_{x=z+1}^N \sum_{y=z+1}^{x-1} \frac{1}{x^a y^b z^c} + \sum_{y=1}^N \sum_{x=y+1}^N \sum_{z=1}^{y-1}\frac{1}{x^a y^c z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\
&\:\:\:\:\:\text{ swapping order of summation on the first sum} \\
&= \sum_{x=1}^N \sum_{y=x+1}^N \sum_{z=x+1}^{y-1} \frac{1}{x^c y^a z^b} + \sum_{x=1}^N \sum_{y=x+1}^N \sum_{z=1}^{x-1}\frac{1}{x^c y^a z^b} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\
&\:\:\:\:\: \text{ relabeling variables on the first two sums} \\
&= \sum_{x=1}^N \sum_{y=x+1}^N \sum_{z=1}^{y-1} \frac{1}{x^c y^a z^b} - \sum_{x=1}^N \sum_{y=x+1}^N \frac{1}{x^{b+c} y^a} + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1}\frac{1}{x^c y^a z^b}, \\
&\:\:\:\:\: \text{ combining the first two sums} \\
&= \sum_{x=1}^N \sum_{y=1}^N \sum_{z=1}^{y-1} \frac{1}{x^c y^a z^b} - \sum_{x=1}^N \sum_{z=1}^{y-1} \frac{1}{x^{a+c} z^b} - \sum_{y=1}^N \sum_{x=1}^{y-1} \frac{1}{x^{b+c} y^a}, \\
&\:\:\:\:\:\text{ combining the first and third sums and swapping the order of summation on the second} \\
&= \zeta_N(c) \zeta_N(a,b) - \zeta_N(a+c,b) - \zeta_N(a,b+c). \square
\end{align}

Proof of (3):
\begin{align}
\zeta_N(2,2,1) &= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2 y^2 z} \\
&= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2 y^2 (y-z)}, \text{ reindexing on the third sum} \\
&= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \left( -\frac{1}{x^2 y z^2} - \frac{1}{x^2 y^2 z} + \frac{1}{x^2(y-z)z^2} \right), \\
&\:\:\:\:\: \text{ by partial fractions decomposition} \\
&= - \zeta_N(2,1,2) - \zeta_N(2,2,1) + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2(y-z)z^2} \tag{5}. \\
\end{align}
Now, let's look at the third expression in (5).

\begin{align}
&\sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{x^2(y-z)z^2} \\
&= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{x-y-1} \frac{1}{x^2(x-y-z)z^2}, \text{ reindexing the second sum} \\
&= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=y+1}^{x-1} \frac{1}{x^2(x-z)(z-y)^2}, \text{ reindexing the third sum} \\
&= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=1}^{z-1} \frac{1}{x^2(x-z)(z-y)^2}, \text{ swapping the order of summation} \\
&= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=1}^{z-1} \frac{1}{x^2(x-z)y^2}, \text{ reindexing the third sum} \\
&= \sum_{x=1}^N \sum_{z=1}^{x-1} \sum_{y=1}^{z-1} \left(-\frac{1}{x y^2 z^2} - \frac{1}{x^2 y^2 z} + \frac{1}{(x-z)y^2 z^2} \right), \text{ by partial fractions decomposition} \\
&= - \zeta_N(1,2,2) - \zeta_N(2,1,2) + \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{(x-y)y^2 z^2} \tag{6}, \text{ relabeling variables}.
\end{align}
Let's look at the third expression in (6).

\begin{align}
&\sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{(x-y)y^2 z^2} \\
&= \sum_{x=1}^N \sum_{y=1}^{x-1} \sum_{z=1}^{y-1} \frac{1}{(x-y)y^2 z^2} + \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2} - \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2} \\
&= \left(\sum_{x=1}^N \frac{1}{x}\right) \left(\sum_{y=1}^N \sum_{z=1}^{y-1} \frac{1}{y^2 z^2} \right) - \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2}, \\
&\:\:\:\:\: \text{ via the finite sum version of the Cauchy product} \\
&= \zeta_N(1) \zeta_N(2,2) - e_N(1,2,2), \tag{7} \\
\end{align}
where
$$e_N(1,2,2) = \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2}.$$
Putting (5), (6), and (7) together, we have

\begin{align}
\zeta_N(2,2,1) =& - \zeta_N(2,1,2) - \zeta_N(2,2,1) - \zeta_N(1,2,2) - \zeta_N(2,1,2) + \zeta_N(1) \zeta_N(2,2) \\
&- e_N(1,2,2), \\
\zeta_N(2,2,1) + \zeta_N(2,1,2) &= - \zeta_N(1) \zeta_N(2,2) + \zeta_N(2,3) + \zeta_N(3,2) + \zeta_N(1) \zeta_N(2,2) \\
&- e_N(1,2,2), \text{ by (4)} \\
=& \zeta_N(2,3) + \zeta_N(3,2) - e_N(1,2,2). \\
\end{align}
All that remains to complete the proof of (3) is to show that $e_N(1,2,2) \to 0$ as $N \to \infty$. We have
\begin{align}
e_N(1,2,2) &= \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^{y-1} \frac{1}{x y^2 z^2} \\

&\leq \sum_{x=1}^N \sum_{y=N+1-x}^N \sum_{z=1}^N \frac{1}{x y^2 z^2} \\
&= \zeta_N(2) \sum_{x=1}^N \sum_{y=N+1-x}^N \frac{1}{x y^2} \\
&= \zeta_N(2) \sum_{y=1}^N \sum_{x=N+1-y}^N \frac{1}{x y^2}, \text{ swapping the order of summation} \\
&\leq \zeta_N(2) \sum_{y=1}^N \frac{1}{y^2} \sum_{x=N+1-y}^N \frac{1}{N+1-y} \\
&= \zeta_N(2) \sum_{y=1}^N \frac{1}{y^2} \frac{y}{N+1-y} \\
&= \zeta_N(2) \sum_{y=1}^N \frac{1}{y (N+1-y)}\\
&= \zeta_N(2) \frac{1}{N+1}\sum_{y=1}^N \left(\frac{1}{y} + \frac{1}{N+1-y} \right), \text{ by partial fractions decomposition} \\
&= \zeta_N(2) \frac{2}{N+1} \zeta_N(1),
\end{align}
which goes to $0$ as $N \to \infty$, since $\zeta_N(1) = O(\log N)$ and $\zeta_N(2) = O(1)$. $\square$

How to simplify this fraction?

Can anyone show me how to simplify this fraction:

$$
\frac{(k + 1)((k + 1)+1)(2(k + 1)+1)}{6}\;\;.
$$

What can be factored out and so forth?

Thanks.

Answer

$$
\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}
$$

Then if $k=2n$ you have

\begin{align*}
\frac{(k+1)(k+2)(2k+3)}{6}=&

\frac{(2n+1)(2n+2)(4n+3)}{6}\\
=&\frac{(2n+1)2(n+1)(4n+3)}{6}\\
=&\frac{(2n+1)(n+1)(4n+3)}{3}
\end{align*}

it's not so different if $k=2n+1$.
Then if you want you can consider in more detail: what if $n=2m$? And $n=2m+1$?
Enjoy!

modular arithmetic - Prove by induction that if $ aequiv b pmod m$ then $a^n equiv b^n pmod m$

The base case is pretty straightforward. But I'm stuck on the inductive step.

As the base case holds, assume for when $n=k$ holds, show the $k+1$ case holds true.

Inductive Hypothesis: $a^k \equiv b^k \pmod m$, then

$$a^{k+1} \equiv b^{k+1} \pmod m \iff a^{k+1} - b^{k+1} = m(k), k \in \mathbb{Z}. $$

I think I'm missing some steps, I'm not sure how to manipulate what I have to shows it holds.

sequences and series - Prove for a triangle whose side are in AP

If the sides of a triangle are in AP and the greatest angle exceeds the smallest angle by a show that the sides are in the ratio 1 - X :1: 1 + X where X = √((1- cos a)(7-cos a))