calculus - Convergence of $int_0 ^{infty}frac{cos t}{t^{alpha}} dt$ related to $Gamma$ function

I would like to show that the integral

$$\int_0 ^\infty \frac{\cos t}{t^{\alpha}} dt$$ converges for $0<\alpha <1$. I already showed that it does not converge for $\alpha\leq 0$ or $\alpha \geq 1$. Do you have any hints for me? I know that I can use the gamme function (using $\cos t=\frac{e^{it}+e^{-it}}{2}$) but I don't see why I can use the gamma function just for $\alpha \in (0,1)$.

Answer

Hint. Recall that, from the definition of the Euler $\Gamma$ function, we have
$$
\begin{align}
\int_{0}^{\infty} e^{-bt} \, t^{-\alpha} \, dt = \frac{\Gamma(1-\alpha)}{b^{1-\alpha}}, \quad 0<\alpha<1, \Re b>0. \tag1
\end{align}
$$ Then put $b:=b_\epsilon:=\epsilon+i,\, \epsilon>0$, in $(1)$, let $\epsilon \to 0^+$ and take the real part to get

$$
\begin{align}
\int_{0}^{\infty} t^{-\alpha} \cos t \, dt & = \sin \left(\frac{\pi \alpha}{2}\right)\Gamma(1-\alpha), \quad 0<\alpha<1. \tag2
\end{align}
$$

Extension of the additive Cauchy functional equation

Let $f\colon (0,\alpha)\to \def\R{\mathbf R}\R$ satisfy $f(x + y)=f(x)+f(y)$

for all $x,y,x + y \in (0,\alpha)$, where $\alpha$ is a positive real number. Show that there exists an additive function $A \colon \R \to \R$ such that $A(x) = f(x)$ for all $x \in (0, \alpha)$.
Simply I want to define a function A In specific form as an extension of the function f wich is additive functional equation. I tried to define the function A .

Answer

Let $x > 0$. Choose $n \in \def\N{\mathbf N}\N$ with $\frac xn < \alpha$. Define $A(x) := nf(\frac xn)$. Note that this is well-defnined: If $m\in \N$ is another natural number such that $\frac xm < \alpha$, we have
\begin{align*}
mf\left(\frac xm\right) &= mf\left(\sum_{k=1}^n \frac x{mn}\right)\\
&= m\sum_{k=1}^n f\left(\frac x{mn}\right)\\
&= \sum_{l=1}^m n f\left(\frac x{mn}\right)\\
&= nf\left(\sum_{l=1}^m \frac x{mn}\right)\\
&= nf\left(\frac x{n}\right).

\end{align*}
For $x < 0$ choose $n \in \N$ with $\frac xn > -\alpha$ and define $A(x) := -nf(-\frac xn)$, finally, let $A(0) = 0$. Then $A$ is an extension of $f$, to show that it is additive, let $x,y \in \def\R{\mathbf R}\R$. Choose $n \in \N$ such that $\frac xn, \frac yn, \frac{x+y}n \in (-\alpha, \alpha)$. We have if $x,y \ge 0$:
\begin{align*}
A(x+y) &= nf\left(\frac{x+y}n\right)\\
&= nf\left(\frac xn\right) + nf\left(\frac yn\right)\\
&= A(x) + A(y)
\end{align*}
If both $x,y \le 0$, we argue along the same lines. Now suppose $x \ge 0$, $y \le 0$, $x+y \ge 0$. We have $A(y) = -A(-y)$ be definition of $A$. Hence
\begin{align*}
-A(y) + A(x+y) &= A(-y) + A(x+y)\\

&= A(-y+x+y)\\
&= A(x).
\end{align*}
If $x \ge 0$, $y \le 0$, $x+y \le 0$, we have $-x \le 0$ and
\begin{align*}
-A(x) + A(x+y) &= A(-x) + A(x+y)\\
&= A(y)
\end{align*}

Progression arithmetic/geometric

I tried to equate the system but could not find the solution. Could anyone help?

The sequence $(a_1, a_2, a_3, \ldots)$ is an arithmetic progression with common difference $3$, and the sequence $(b_1, b_2, b_3, \ldots)$ is an increasing geometric progression. Knowing that $a_2 = b_3$, $a_{10} = b_5$ and $a_{42} = b_7$, the value of $b_4-a_4$ is:

Given Options

• a. 2




• b. 0




• c. 1




• d. -1

number theory - Can we prove that there are no two perfect powers with difference $6$?

Here :

Are those lists known to be complete?

I asked whether the list of numbers given in the link is known to be complete. Since the generalized Catalan-conjecture is apparantly open, I think they are not. In particular I am interested in the smallest candidate, for which there might be no solution, the case $n=6$

Can we proof that there are no two perfect powers with difference $6$ ? If not, can we at least prove that such powers must be very large ?

It is clear that two perfect powers with difference $6$ have the same parity. Furthermore, the powers cannot be even because then both powers would be divisble by $4$, therefore the difference would be divisble by $4$ as well, which is impossible because $4$ does not divide $6$.

We also can easily see that $6$ cannot be the difference of two squares. And apparantly, the Mordell-curves $y^2=x^3+6$ and $y^2=x^3-6$ both have no integral solutions either.

Answer

There is no integer $c$ with absolute value exceeding $1$ for which we can even prove that the number of solutions to the equation
$$
x^n-y^m=c
$$

is finite (though such a result would follow trivially from the ABC-conjecture). The case with $c=6$ is indeed the smallest positive value where we expect there to be no solutions (which follows from a suitably explicit version of ABC), but, other than being able to show that there are no solutions to $x^2-y^n = \pm 6$ (and, presumably, being able to handle a few ``small'' pairs $m, n$), I don't believe there is anything more that we can prove with current technology.

The equations $x^2+6 = y^n$ and $x^2-6=y^n$ were solved by J. H. E.Cohn and by C. F. Barros (a student of Samir Siksek), respectively. The first of these is a relatively elementary argument (the paper is in Acta Arithmetic, from 1993) while the second is not.

The longest repeating decimal that can be created from a simple fraction

What's the longest possible repeating decimal (repetend) that can be created from a fraction if:

• The numerator has to be less than or equal to 9,999


• The denominator has be less than or equal to 9,999?

I know the repeating decimal part can't exceed the denominator - 1. So the longest possible repeating decimal part has to be 9,998 or less.

The reason I want to know is to test an algorithm that I wrote which accepts fractions with numerators and denominators up to 9,999. The largest repeating decimal part I was able to create so far was 1/97 which equaled 0.[01030927 83505154 63917525 77319587 62886597 93814432 98969072 16494845 36082474 22680412 37113402 06185567] (96 repeating digits).

Answer

The numerator doesn't matter (for this question), so you might as well let it be $1$. The denominator should be the largest prime under $10000$ which has $10$ as a primitive root. I don't know offhand what that prime is, but I'm sure such primes are tabulated and shouldn't be hard to locate.

The table at the Online Encyclopedia doesn't go far enough. There is an applet which claims to find these primes, but I couldn't make it work --- maybe you'll have better luck.

calculus - Find: $lim_{n to infty} int_0^{infty} arctan(nx) e^{- x^n}dx$

Find:

$$\lim_{n \to \infty} \int_0^{\infty} \arctan(nx) e^{- x^n}dx$$

Probably, no recursive form could be found, and elementary tools (integration by parts, change of variable, etc.) are not useful here. How can I find such a limit?

Thank you.

Answer

Start by thinking about pointwise limits. For $x>0$, $\arctan(nx) \to \pi/2$. For $01$, $x^n \to +\infty$. Hence for $01$, $e^{-x^n} \to 0$. So the pointwise limit is $\frac{\pi}{2} \chi_{(0,1)}$, except maybe at $0$ and $1$ which don't matter.

So we might intuitively expect the limit to be $\int_0^1 \frac{\pi}{2} dx = \frac{\pi}{2}$. Try to use an integral convergence theorem, such as the dominated convergence theorem, to justify this result.

sequences and series - To test the convergence $sum{frac{1}{(ln {n})^{ln{n}}}}$

How to test the convergence of following series
$$\sum{\frac{1}{(\ln {n})^{\ln{n}}}}$$

I have tried Cauchy condensation test and gives me nothing