sequences and series - Does $sumlimits_{k=1}^n 1 / k ^ 2$ converge when $nrightarrowinfty$?

I can prove this sum has a constant upper bound like this:

$$\sum_{k=1}^n \frac1{k ^ 2} \lt 1 + \sum_{k=2}^n \frac 1 {k (k - 1)} = 2 - \frac 1 n \lt 2$$

And computer calculation shows that sum seems to converge to 1.6449. But I still want to know:

• Dose this sum converge?


• Is there a name of this sum (or the series $1 / k ^2 $)?

Answer

A sequence that is increasing and bounded must converge. That's one of the fundamental properties of the real line. So once you've observed that your sequence of partial sums is bounded, since it obviously increases, it must converge. Of course it is a very famous series, and it converges to a number which quite miraculously has a "closed form" formula: it is $\pi^2/6$.

EDIT: for many proofs of this famous formula, see this MO question.

special functions - Extending the result $int_{0}^{infty} left( ( 1 - 2C(x))^{2} + (1-2S(x))^{2} right) , dx = frac{4}{pi} $

While generalizing this result, I succeeded in proving that for $\alpha > 0$, $\beta < 1$ and $1 < 2\alpha + \beta < 3$, we have

\begin{align*}
&\int_{0}^{\infty} \left[ \left( \int_{x}^{\infty} \frac{\cos t}{t^{\alpha}} \, dt \right)^{2} + \left( \int_{x}^{\infty} \frac{\sin t}{t^{\alpha}} \, dt \right)^{2} \right] \, \frac{dx}{x^{\beta}} \\
& \hspace{10em} = \frac{\pi}{1-\beta} \frac{\Gamma(2-\alpha-\beta)}{\Gamma(\alpha)} \csc \left( \pi \alpha + \frac{\pi (\beta-1)}{2} \right).
\end{align*}

My question is

1. Is this a known result?


2. My ultimate goal is to examine whether the integral
   $$ I(\alpha, \beta) := \int_{0}^{\infty} \left[ \left( \int_{x}^{\infty} \frac{\cos t}{t^{\alpha}} \, dt \right)^{2} + \left( \int_{x}^{\infty} \frac{\sin t}{t^{\alpha}} \, dt \right)^{2} \right]^{2} \, \frac{dx}{x^{\beta}} $$
   has closed from or not for general $\alpha$ and $\beta$. I know that
   $$ I \left(\tfrac{1}{2}, 0 \right) = 2\pi (\log 4 - 1) \qquad \text{and} \qquad I \left(1, 0 \right) = \frac{2\pi^{3}}{3}, \tag{1} $$
   but I know nothing for the other cases. (Here, the former identity in $(1)$ corresponds to the motivating problem linked above.) Is there any other known result concerning this integral?

---

A further inspection showed that

\begin{align*}
\int_{0}^{\infty} \left( ( 1 - 2C(x))^{2} + (1-2S(x))^{2} \right)^{2} \, dx
&= \frac{16}{\pi} \left( 1 - \frac{2\sqrt{2}}{\pi} \log \left( 1 + \sqrt{2} \right) \right) \\
&\approx 1.0516193625061961290 \cdots,
\end{align*}

where

$$ C(x) = \int_{0}^{x} \cos \left( \tfrac{\pi t^2}{2} \right) \, dt \quad \text{and} \quad S(x) = \int_{0}^{x} \sin \left( \tfrac{\pi t^2}{2} \right) \, dt $$

are Fresnel integrals. Indeed, this corresponds to

$$ I \left( \tfrac{1}{2}, \tfrac{1}{2} \right) = \sqrt{\pi} \left( 4 \sqrt{2} \pi - 16 \log \left( 1 + \sqrt{2} \right) \right). $$

Note that major inverse symbolic calculators do not yield this result.

Answer

Pardon My Progress...

First, the sum of squared integrals inside the brackets, which for convenience we shall denote by $f{\left(x;\alpha\right)}$, may be rewritten as a single double integral with a little algebra and trigonometry:

$$\begin{align}
f{\left(x;\alpha\right)}
&:=\left(\int_{x}^{\infty}\frac{\cos{t}}{t^\alpha}\,\mathrm{d}t\right)^2+\left(\int_{x}^{\infty}\frac{\sin{t}}{t^\alpha}\,\mathrm{d}t\right)^2\\
&=\small{\left(\int_{x}^{\infty}\frac{\cos{t_1}}{t_1^\alpha}\,\mathrm{d}t_{1}\right)\cdot\left(\int_{x}^{\infty}\frac{\cos{t_2}}{t_2^\alpha}\,\mathrm{d}t_{2}\right)+\left(\int_{x}^{\infty}\frac{\sin{t_1}}{t_1^\alpha}\,\mathrm{d}t_{1}\right)\cdot\left(\int_{x}^{\infty}\frac{\sin{t_2}}{t_2^\alpha}\,\mathrm{d}t_{2}\right)}\\
&=\int_{x}^{\infty}\int_{x}^{\infty}\left(\frac{\cos{t_1}}{t_1^\alpha}\cdot\frac{\cos{t_2}}{t_2^\alpha}\right)\,\mathrm{d}t_{1}\mathrm{d}t_{2}+\int_{x}^{\infty}\int_{x}^{\infty}\left(\frac{\sin{t_1}}{t_1^\alpha}\cdot\frac{\sin{t_2}}{t_2^\alpha}\right)\,\mathrm{d}t_{1}\mathrm{d}t_{2}\\

&=\int_{x}^{\infty}\int_{x}^{\infty}\left(\frac{\cos{\left(t_1\right)}\cos{\left(t_2\right)}+\sin{\left(t_1\right)}\sin{\left(t_2\right)}}{\left(t_{1}t_{2}\right)^\alpha}\right)\,\mathrm{d}t_{1}\mathrm{d}t_{2}\\
&=\int_{x}^{\infty}\int_{x}^{\infty}\frac{\cos{\left(t_1-t_{2}\right)}}{\left(t_{1}t_{2}\right)^\alpha}\,\mathrm{d}t_{1}\mathrm{d}t_{2}.\\
\end{align}$$

Next, we apply a sequence of two two-variable transformations to put the integral into a more tractable form. The first substitution is a simple scaling transformation, $(1)$$(t_{1},t_{2})=(xu_{1},xu_{2})$; the second substitution is the more complicated transformation, $(2)$ $(u_{1}-u_{2},u_{1}u_{2})=(w_{1},w_{2})$:

$$\begin{align}
\mathcal{I}{\left(\alpha,\beta\right)}
&=\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta}}\,\left[\left(\int_{x}^{\infty}\frac{\cos{t}}{t^\alpha}\,\mathrm{d}t\right)^2+\left(\int_{x}^{\infty}\frac{\sin{t}}{t^\alpha}\,\mathrm{d}t\right)^2\right]^2\\
&=\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta}}\,\left[f{\left(x;\alpha\right)}\right]^2\\

&=\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta}}\,\left[\int_{x}^{\infty}\int_{x}^{\infty}\frac{\cos{\left(t_1-t_{2}\right)}}{\left(t_{1}t_{2}\right)^\alpha}\,\mathrm{d}t_{1}\mathrm{d}t_{2}\right]^2\\
&=\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta}}\,\left[\int_{1}^{\infty}\int_{1}^{\infty}\frac{\cos{\left[x\left(u_1-u_{2}\right)\right]}}{\left(x^2u_{1}u_{2}\right)^\alpha}\,x^2\,\mathrm{d}u_{1}\mathrm{d}u_{2}\right]^2\tag{1}\\
&=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{1}^{\infty}\int_{1}^{\infty}\frac{\cos{\left[x\left(u_1-u_{2}\right)\right]}}{\left(u_{1}u_{2}\right)^\alpha}\,\mathrm{d}u_{1}\mathrm{d}u_{2}\right]^2\\
&=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{1}^{\infty}\mathrm{d}w_{2}\int_{1-w_{2}}^{w_{2}-1}\mathrm{d}w_{1}\,\frac{w_{2}^{-\alpha}\cos{\left(x\,w_{1}\right)}}{\sqrt{w_{1}^2+4w_{2}}}\right]^2\tag{2}\\
&=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{-\infty}^{\infty}\mathrm{d}w_{1}\int_{1+|w_{1}|}^{\infty}\mathrm{d}w_{2}\,\frac{w_{2}^{-\alpha}\cos{\left(x\,w_{1}\right)}}{\sqrt{w_{1}^2+4w_{2}}}\right]^2\\
&=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[2\int_{0}^{\infty}\mathrm{d}w_{1}\int_{1+w_{1}}^{\infty}\mathrm{d}w_{2}\,\frac{w_{2}^{-\alpha}\cos{\left(x\,w_{1}\right)}}{\sqrt{w_{1}^2+4w_{2}}}\right]^2\\
&=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}\mathrm{d}w_{1}\int_{1+w_{1}}^{\infty}\mathrm{d}w_{2}\,\frac{w_{2}^{-\alpha}\cos{\left(x\,w_{1}\right)}}{\sqrt{\frac14w_{1}^2+w_{2}}}\right]^2\\
&=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}\mathrm{d}w_{1}\,\cos{\left(x\,w_{1}\right)}\int_{1+w_{1}}^{\infty}\,\frac{w_{2}^{-\alpha}\,\mathrm{d}w_{2}}{\sqrt{\frac14w_{1}^2+w_{2}}}\right]^2.\\
\end{align}$$

To perform the integration with respect to $w_{2}$, I appeal to Gradshteyn 3.197(2): under the conditions $|\arg{\frac{u}{\beta}}|<\pi\lor |\frac{\beta}{u}|<1$, and $0<\Re{\left(\mu\right)}<\Re{\left(\lambda-\nu\right)}$, we have the result,

$$\int_{u}^{\infty}x^{-\lambda}(x+\beta)^{\nu}(x-u)^{\mu-1}\,\mathrm{d}x=u^{-\lambda}(\beta+u)^{\mu+\nu}\operatorname{B}{\left(\lambda-\mu-\nu,\mu\right)}\times{_2F_1}{\left(\lambda,\mu;\lambda-\mu;-\frac{\beta}{u}\right)}.$$

Let $\mu=1$, $\nu=-\frac12$, $\lambda=\alpha$, $u=1+w_{1}$, and $\beta=\frac14w_{1}^2$. Then for $\frac12<\Re{\left(\alpha\right)}<1\land w_{1}>0$,

$$\begin{align}
\int_{1+w_{1}}^{\infty}\frac{w_{2}^{-\alpha}\,\mathrm{d}w_{2}}{\sqrt{\frac14w_{1}^2+w_{2}}}

&=\small{\left(w_{1}+1\right)^{-\alpha}\sqrt{\frac14w_{1}^2+w_{1}+1}\,\operatorname{B}{\left(\alpha-\frac12,1\right)}\,{_2F_1}{\left(\alpha,1;\alpha-1;-\frac{\frac14w_{1}^2}{1+w_{1}}\right)}}\\
&=\left(w_{1}+1\right)^{-\alpha}\left(\frac{w_{1}+2}{2}\right)\,\left(\frac{2}{2\alpha-1}\right)\,{_2F_1}{\left(\alpha,1;\alpha-1;-\frac{w_{1}^2}{4\left(1+w_{1}\right)}\right)}\\
&=\frac{\left(w_{1}+2\right)\left(w_{1}+1\right)^{-\alpha}}{2\alpha-1}\,{_2F_1}{\left(\alpha,1;\alpha-1;-\frac{w_{1}^2}{4\left(1+w_{1}\right)}\right)}.\\
\end{align}$$

The hypergeometric function above quite conveniently reduces to a rational function for the specified combination of parameters: for $\alpha\neq1$,

$${_2F_1}{\left(\alpha,1;\alpha-1;-z\right)}=\frac{\alpha\,z+\alpha-2z-1}{(\alpha-1)(z+1)^2},\\
\implies {_2F_1}{\left(\alpha,1;\alpha-1;-\frac{w_{1}^2}{4\left(1+w_{1}\right)}\right)}=\frac{4\left(w_{1}+1\right)\left[\alpha(w_{1}+2)^2-2(w_{1}+1)^2-2\right]}{(\alpha-1)(w_{1}+2)^4}.\\$$

Thus, for $\frac12<\Re{\left(\alpha\right)}<1\land w_{1}>0$,

$$\begin{align}
\int_{1+w_{1}}^{\infty}\frac{w_{2}^{-\alpha}\,\mathrm{d}w_{2}}{\sqrt{\frac14w_{1}^2+w_{2}}}
&=\frac{\left(w_{1}+2\right)\left(w_{1}+1\right)^{-\alpha}}{2\alpha-1}\,{_2F_1}{\left(\alpha,1;\alpha-1;-\frac{w_{1}^2}{4\left(1+w_{1}\right)}\right)}\\
&=\frac{4\left(w_{1}+1\right)^{1-\alpha}\left[\alpha(w_{1}+2)^2-2(w_{1}+1)^2-2\right]}{(2\alpha-1)(\alpha-1)(w_{1}+2)^3},\\
\end{align}$$

and hence:

$$\begin{align}
\mathcal{I}{\left(\alpha,\beta\right)}
&=\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}\mathrm{d}w_{1}\,\cos{\left(x\,w_{1}\right)}\int_{1+w_{1}}^{\infty}\,\frac{w_{2}^{-\alpha}\,\mathrm{d}w_{2}}{\sqrt{\frac14w_{1}^2+w_{2}}}\right]^2\\
&=\small{\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta-4\left(1-\alpha\right)}}\left[\int_{0}^{\infty}\frac{4\left(w_{1}+1\right)^{1-\alpha}\left[\alpha(w_{1}+2)^2-2(w_{1}+1)^2-2\right]\cos{\left(xw_{1}\right)}}{(2\alpha-1)(\alpha-1)(w_{1}+2)^3}\,\mathrm{d}w_{1}\right]^2}\\
&=\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta-4\left(1-\alpha\right)}}\left[\int_{0}^{\infty}\frac{4\left(y+1\right)^{1-\alpha}\left[\alpha(y+2)^2-2(y+1)^2-2\right]\cos{\left(xy\right)}}{(2\alpha-1)(\alpha-1)(y+2)^3}\,\mathrm{d}y\right]^2\\
&=\small{\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\beta-4\left(1-\alpha\right)}}\left[\int_{0}^{\infty}\frac{\left(y+1\right)^{1-\alpha}\left[\alpha(y+2)^2-2(y+1)^2-2\right]\cos{\left(xy\right)}}{(y+2)^3}\,\mathrm{d}y\right]^2},\\
&=:\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}g{\left(y;\alpha\right)}\,\cos{\left(xy\right)}\,\mathrm{d}y\right]^2,\\
\end{align}$$

where in the third line we've made the substitution $w_{1}=y$ for the sake of eliminating the dependency on subscripted variables, and where in the last line we've introduced the auxiliary function $g{\left(y;\alpha\right)}$ simply for the sake conveniently denoting the function,

$$\begin{align}
g{\left(y;\alpha\right)}
&:=\frac{\left(y+1\right)^{1-\alpha}\left[\alpha(y+2)^2-2(y+1)^2-2\right]}{(y+2)^3}\\
&=\frac{\alpha\left(y+1\right)^{1-\alpha}}{y+2}-\frac{2\left(y+1\right)^{3-\alpha}}{(y+2)^3}-\frac{2\left(y+1\right)^{1-\alpha}}{(y+2)^3}.\\
\end{align}$$

Now if we repeat our initial trick of rewriting the square of an integral as a double integral, we can arrive at an expression for $\mathcal{I}{\left(\alpha,\beta\right)}$ as an ordinary triple integral:

$$\begin{align}

\mathcal{I}{\left(\alpha,\beta\right)}
&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}g{\left(y;\alpha\right)}\,\cos{\left(xy\right)}\,\mathrm{d}y\right]^2\\
&=\small{\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}g{\left(y;\alpha\right)}\,\cos{\left(xy\right)}\,\mathrm{d}y\right]\cdot\left[\int_{0}^{\infty}g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\,\mathrm{d}z\right]}\\
&=\small{\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\,x^{4\left(1-\alpha\right)-\beta}\left[\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}\right]}\\
&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,\frac{g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}}{x^{\beta-4\left(1-\alpha\right)}}.\\
\end{align}$$

Change the order of integration so that the integration over $x$ is first:

$$\begin{align}

\mathcal{I}{\left(\alpha,\beta\right)}
&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\frac{g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}}{x^{\beta-4\left(1-\alpha\right)}}\\
&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\int_{0}^{\infty}\mathrm{d}x\frac{g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\,\cos{\left(xz\right)}\cos{\left(xy\right)}}{x^{\beta-4\left(1-\alpha\right)}}\\
&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\int_{0}^{\infty}\mathrm{d}x\frac{\cos{\left(xy\right)}\cos{\left(xz\right)}}{x^{\beta-4\left(1-\alpha\right)}}.\\
\end{align}$$

Then the inner integral with respect to $x$ may be evaluated in closed form with an appeal to another result from Gradshteyn. Proposition 3.762(3) of Gradshteyn states that, given $a,b\in\mathbb{R}$ and $\mu\in\mathbb{C}$ such that $a>0,~b>0$, and $0<\Re{(\mu)}<1$, then the following improper integral has the closed form:

$$\int_{0}^{\infty}x^{\mu-1}\cos{\left(ax\right)}\cos{\left(bx\right)}\,\mathrm{d}x=\frac12\cos{\left(\frac{\mu\pi}{2}\right)}\,\Gamma{\left(\mu\right)}\,\left[\left(a+b\right)^{-\mu}+|a-b|^{-\mu}\right].$$

Setting $(a,b,\mu)\mapsto(y,z,4(1-\alpha)-\beta+1)$ in the above proposition yields the following corrollary: given $y,z\in\mathbb{R}$ and $p\in\mathbb{C}$ such that $y>0,~z>0$, and $-1<\Re{\left(4(1-\alpha)-\beta\right)}<0$, then the following improper integral has the closed form,

$$\small{\int_{0}^{\infty}\frac{\cos{\left(xy\right)}\cos{\left(xz\right)}}{x^{\beta+4\alpha-4}}\mathrm{d}x=\frac12\sin{\left[\frac{\pi\left(4\alpha+\beta\right)}{2}\right]}\,\Gamma{\left(5-4\alpha-\beta\right)}\,\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right].}$$

Thus, we may reduce our integral representation of $\mathcal{I}{\left(\alpha,\beta\right)}$ to a single double integral:

$$\begin{align}
\mathcal{I}{\left(\alpha,\beta\right)}

&=\frac{16}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\int_{0}^{\infty}\mathrm{d}x\frac{\cos{\left(xy\right)}\cos{\left(xz\right)}}{x^{\beta-4\left(1-\alpha\right)}}\\
&=\small{\frac{8\sin{\left[\frac{\pi\left(4\alpha+\beta\right)}{2}\right]}\,\Gamma{\left(5-4\alpha-\beta\right)}}{(2\alpha-1)^2(\alpha-1)^2}\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]}\\
&=:\frac{8\sin{\left[\frac{\pi\left(4\alpha+\beta\right)}{2}\right]}\,\Gamma{\left(5-4\alpha-\beta\right)}}{(2\alpha-1)^2(\alpha-1)^2}\tilde{\mathcal{I}}{\left(\alpha,\beta\right)}.\\
\end{align}$$

---

Update:

Now we'll focus on reducing the previously defined function $\tilde{\mathcal{I}}{\left(\alpha,\beta\right)}$. First of all, by symmetry we can reduce the region of integration to one where the absolute value bars are no longer necessary inside the integrand, which will obviate some of tedium of evaluation:

$$\begin{align}
\tilde{\mathcal{I}}{\left(\alpha,\beta\right)}
&=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&~~~~~ +\int_{0}^{\infty}\mathrm{d}y\int_{y}^{\infty}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&~~~~~ +\int_{0}^{\infty}\mathrm{d}z\int_{0}^{z}\mathrm{d}y\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&~~~~~ +\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(z;\alpha\right)}\,g{\left(y;\alpha\right)}\left[\left(z+y\right)^{\beta+4\alpha-5}+|z-y|^{\beta+4\alpha-5}\right]\\

&=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&~~~~~ +\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+|y-z|^{\beta+4\alpha-5}\right]\\
&=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+\left(y-z\right)^{\beta+4\alpha-5}\right].\\
\end{align}$$

Next, we rescale the interval of integration of the inner integral to the unit interval via the substitution $z=y\,\omega$:

$$\begin{align}
\tilde{\mathcal{I}}{\left(\alpha,\beta\right)}

&=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,g{\left(y;\alpha\right)}\,g{\left(z;\alpha\right)}\left[\left(y+z\right)^{\beta+4\alpha-5}+\left(y-z\right)^{\beta+4\alpha-5}\right]\\
&=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{1}y\,\mathrm{d}\omega\,g{\left(y;\alpha\right)}\,g{\left(y\,\omega;\alpha\right)}\left[\left(y+y\,\omega\right)^{\beta+4\alpha-5}+\left(y-y\,\omega\right)^{\beta+4\alpha-5}\right]\\
&=2\int_{0}^{\infty}\mathrm{d}y\int_{0}^{1}y^{\beta+4\alpha-4}\,\mathrm{d}\omega\,g{\left(y;\alpha\right)}\,g{\left(y\,\omega;\alpha\right)}\left[\left(1+\omega\right)^{\beta+4\alpha-5}+\left(1-\omega\right)^{\beta+4\alpha-5}\right]\\
&=2\int_{0}^{\infty}\mathrm{d}y\,y^{\beta+4\alpha-4}\,g{\left(y;\alpha\right)}\int_{0}^{1}\mathrm{d}\omega\,g{\left(y\,\omega;\alpha\right)}\left[\left(1+\omega\right)^{\beta+4\alpha-5}+\left(1-\omega\right)^{\beta+4\alpha-5}\right].\\
\end{align}$$

At this stage, the inner integral over $\omega$ may be evaluated in terms of the Appell hypergeometric function, $F_{1}$.

real analysis - $f(nx)to 0$ as $nto+infty$

Let $f:\mathbb R^+\to\mathbb R^+$ be a continuous function, and let $I$ be a subset of $\mathbb R^+$ such that the following property holds:

For any $x\in I$, $f(nx)\to 0$ as $n\to+\infty$.

Intuitively, if $I$ is 'big' enough, $f$ necessarily tends to $0$ at infinity, but it happens to not always be the case. I am investigating whether it can be said, for various $I$, if

$f(x)\to 0$ as $x\to+\infty$.

---

As a first example, consider $I=[0,1]$. Set $\varepsilon>0$, and consider the closed sets $F_n=\{x\in I\mid f(kx)<\varepsilon,\forall k\geq n\}$. Their union is $[0,1]$, and thus, thanks to the property of Baire, one of them has non-empty interior, i.e. $[a,b]\in F_n$ for a certain $n$. It follows that for all $k\geq n$ and $x\in[ka,kb]$, $f(x)<\varepsilon$. But since $b>a$, $\bigcup_{k=n}^\infty [ka,kb]$ contains a half-line, and $\limsup_{x\to\infty} f(x) \leq \varepsilon$. Since the reasoning holds for any $\varepsilon>0$, we conclude that $f$ does, indeed, tend to $0$ at infinity. The same proof actually works for all $I$ with non-empty interior.

The result is false in higher dimensions when $I$ contains a neighbourhood of the origin, as a simple counter-example can be constructed using a parabola.

What happens when $I\subset \mathbb R$ is smaller? Consider the following sets:

1. $I$ is any measurable set with empty interior, but with Lebesgue measure >0. In that case, one of the aforementioned $F_n$ has positive Lebesgue measure.




2. $I=(0,1)\cap C$, where $C$ is the Cantor set (or another uncountable set). Then, one of the $F_n$ has to be uncountable aswell,




3. $I=\{1/k,k\in\mathbb N\}$ or $I=(0,1)\cap \mathbb Q$, both of these being equivalent. I provided an answer below for this one, and actually for any countable set; such a function $f$ does not necessarily converge to $0$.

In any of these cases, can anything be said about the behaviour of $f$ at infinity?

Bonus question: what is the minimum condition for $I$ (if there is one) so that $f$ has to converge to $0$?

---

I thought about mimicking the proof of the first example when $I$ has empty interior but positive Lebesgue measure. If there exists an integer $n$ and a non-trivial interval $[a,b]$ such that $|F_n\cap [a,b]|=b-a$, then $f\leq\varepsilon$ on a set that is dense on a half-line, and thus, using continuity, tends to $0$. Sadly, such an interval may not exist, since the closure of $F_n$ could very well be of empty interior, like a generalized Cantor set.

Answer

At the first I remark that your proof works for all non-meager $I$ (that is which are not a countable union of nowhere dense sets). From the other hand, let $I$ be a meager subset of $\mathbb R^+$. Choose a sequence $\{F_n\}$ of closed nowhere dense sets such that $F_n\subset [1/n;n]$ for each $n$ and $\bigcup F_n\cup \{0\}\supset I$. The family $\mathcal F=\{mF_n:m\ge n^2\}$ is locally finite, so a set $F=\bigcup\mathcal F$ is closed. Since the set $F$ is meager, by Baire theorem, it does not contain a half-line. This means there exists a sequence $S=\{x_n\}\subset\mathbb R^+\setminus\{0\}$ which goes to infinity. Thus the set $S$ is closed. Since the closed sets $F\cup\{0\}$ and $S$ are disjoint and the space $\mathbb R^+$ is normal, there exists a continuous function $f: \mathbb R^+\to \mathbb R^+$ such that $f(F\cup\{0\})=0$ and $f(S)=1$. Therefore there exists arbitrary large $x$ such that $f(x)=1$, but for each $x\in I$ $f(nx)$ eventually gets $0$.

trigonometry - How do I get $cos{theta} lt frac{sin{theta}}{theta} lt 1$?

How do I get:

$$\cos{\theta} \lt \frac{\sin{\theta}}{\theta} \lt 1$$

Answer

For $0< t<\pi/2$:

$\ \ \ \bullet$ Using similar triangles:
$$\color{darkgreen}{\tan t}={\color{maroon}{\sin t}\over\color{darkblue}{\cos t}} ={
{\text{length}( \color{darkgreen} {\overline{{IZ}})} }\over 1 }\quad \Longrightarrow \quad\color{darkgreen}{\tan t}=\text{length}(\color{darkgreen}{\overline{IZ}})$$

$\ \ \ \bullet$ $t$ is the length of the arc $\color{orange}{IQ}$.

$\ \ \ \bullet$ Area of the circular sector $O\color{orange}{IQ}={t\over 2\pi}\cdot \pi\cdot 1^2={t\over2}$.

$\ \ \ \bullet$ Area of $\triangle OQI={1\over2}\cdot1\cdot\color{maroon}{\sin t}$.

$\ \ \ \bullet$ Area of $\triangle OIZ={1\over2}\cdot1\cdot\color{darkgreen}{\tan t}$.

From the diagram we have

$$
\text {area}(\triangle OQI) \le
\text {area}(\text{circular sector} OQI) \le
\text {area}(\triangle OZI)
$$
$$
{1\over2}\cdot1\cdot\sin t\lt{1\over2} t\lt {1\over2}\cdot1\cdot\tan t
$$

$$

\sin t\lt t\lt \cdot\tan t
$$

$$
{1\over\sin t}\gt {1\over t}\gt {\cos t\over \sin t }
$$

$$
\cos t\lt {\sin t\over t}\lt 1.
$$

linear algebra - Linearly Independent Real Numbers over $mathbb{Q}$

Let $n$ be an integer greater than $1$ and let $\sqrt[n]{2}\in \mathbb{R}$ be the unique positive $n$-th root of $2$. Show that the real numbers $1, (\sqrt[n]{2})^2,\cdot\cdot\cdot,(\sqrt[n]{2})^{n-1}$ are linearly independent over $\mathbb{Q}$.

Using the definition of linear independence, the problem is equivalent to showing that
$$c_1+c_2 (\sqrt[n]{2})^2+\cdot\cdot\cdot+c_{n-1}(\sqrt[n]{2})^{n-1}=0$$

has only the trivial solution $c_1=c_2=\cdot\cdot\cdot=c_{n-1}=0$

I believe there is something special about each $\sqrt[n]{2}$ being a unique positive $n$-th root of 2 that allows us to state linear independence. In particular, this implies that each $(\sqrt[n]{2})^i$ is linearly independent where $i\in \{0,2,3,\cdot\cdot\cdot,n-1\}$.

I don't know if my intuition is correct and how to prove this linear independence rigorously.

Answer

Hint: if there were a linear dependence relation between these elements, then $\sqrt[n]{2}$ would satisfy a (monic) polynomial of degree $\leqslant n-1$ (after dividing by the highest nonzero coefficient). On the other hand, $\sqrt[n]{2}$ is a root of $X^{n} - 2$, which is irreducible over $\mathbb{Q}$ by Eisenstein at $2$. Why is this first bit in contradiction with the second fact?

real analysis - Continuous function with non-negative derivative a.e. implies non-decreasing?

Let $f \colon [a,b] \rightarrow \mathbb{R}$ be a continuous function on a compact interval of the real line. Suppose that $f$ is differentiable almost everywhere and that $f'(x) \geq 0$ at every point of differentiability. Is it true that $f$ is non-decreasing on [a,b]?

(If $f$ is $\textit{absolutely}$ continuous, this is certainly true, but I'm not so sure what happens if you weaken the assumption to mere continuity.)

Answer

No. The Cantor-Lebesgue function $f$ is continuous, non-decreasing, non-constant, and satisfies $f'=0$ almost everywhere. It's not hard to see that $f$ is non-differentiable at every point of the Cantor set. So $g=-f$ is a counterexample to your question: $g$ is continuous, differentiable almost everywhere, satisfies $g'\ge 0$ at every point of differentiability, but $g$ is not non-decreasing.

trigonometry - How can I simplify $frac{sin(1) + sin(2) + cdots + sin(100)}{cos(1) + cos(2) + cdots +cos(100)}$

So I was asked to simplify the expression
$$\frac{\sin(1) + \sin(2) + \cdots + \sin(100)}{\cos(1) + \cos(2) + \cdots + \cos(100)}.$$ I'm struggling to find a way of doing it.

I'd like just hints rather than a whole solution if possible.

Thanks in advance