Use Euler's summation formula to prove the following....

[Math100]

Homework Statement

Use Euler's summation formula to prove that, for $x\geq 2$,
$\sum_{n\leq x}\frac{ \log {n}}{n^3}=A-\frac{ \log {x}}{2x^2}-\frac{1}{4x^2}+O\frac{ \log {x}}{x^3}$, where $A$ is a constant.

Relevant Equations

None.

Proof:

Let $x\geq 2$.
Then $\frac{d}{dt}(\frac{ \log {t}}{t^3})=\frac{1-3\log {t}}{t^4}$.
By Euler's summation formula, we have that
$\sum_{n\leq x}\frac{ \log {n}}{n^3}=\int_{1}^{x} \frac{\log {t}}{t^3}dt+\int_{1}^{x} (t-[t])(\frac{1-3\log {t}}{t^4})dt+(x-[x])\frac{log {x}}{x^3}$
$=\frac{-\log {x}}{2x^2}-\frac{1}{4x^2}+\frac{1}{4}+(\int_{1}^{\infty}-\int_{x}^{\infty})(t-[t])\frac{1-3\log {t}}{t^4}dt+O(\frac{log {x}}{x^3})$.
Thus
\begin{align*}
&\left | \int_{x}^{\infty}(t-[t])\frac{1-3\log {t}}{t^4}dt \right |\leq 2\int_{x}^{\infty}\frac{\log {t}}{t^4}dt\\
&=2\cdot (\frac{1+3\log {x}}{9t^3})\\
&=O(\frac{\log {x}}{x^3}).\\
\end{align*}
Therefore, $\sum_{n\leq x}\frac{\log {n}}{n^3}=A-\frac{\log {x}}{2x^2}-\frac{1}{4x^2}+O(\frac{\log {x}}{x^3})$, where $A$ is a constant.

 

[fresh_42]

This needs a lot more explanation. Which formula did you use? I see

What I do not see is how this formula is used to get your step.

 

[Math100]

This needs a lot more explanation. Which formula did you use? I see

View attachment 314701

What I do not see is how this formula is used to get your step.

Euler's summation formula:
If $f$ has a continuous derivative $f'$ on the interval $[y, x]$, where $0<y<x$, then
$\sum_{y<n\leq x}f(n)=\int_{y}^{x}f(t)dt+\int_{y}^{x}(t-[t])f'(t)dt+f(x)([x]-x)-f(y)([y]-y)$.

The formula which you see is something I've never seen before.

 

[fresh_42]

Euler's summation formula:
If $f$ has a continuous derivative $f'$ on the interval $[y, x]$, where $0<y<x$, then
$\sum_{y<n\leq x}f(n)=\int_{y}^{x}f(t)dt+\int_{y}^{x}(t-[t])f'(t)dt+f(x)([x]-x)-f(y)([y]-y)$.

The formula which you see is something I've never seen before.

O.k., let us assume this formula is correct, then you need to show

a) $\displaystyle{ \int_1^\infty (t-[t])\left(\dfrac{1-3\log {t}}{t^4}\right)dt = O\left(\dfrac{\log x}{x^3}\right)}\\[30pt]$
b) $\displaystyle{\left| \int_x^\infty (t-[t])\left(\dfrac{1-3\log {t}}{t^4}\right)dt\right|\leq 2 \int_x^\infty \dfrac{\log t}{t^4}\,dt}$

 

[Math100]

O.k., let us assume this formula is correct, then you need to show

a) $\displaystyle{ \int_1^\infty (t-[t])\left(\dfrac{1-3\log {t}}{t^4}\right)dt = O\left(\dfrac{\log x}{x^3}\right)}\\[30pt]$
b) $\displaystyle{\left| \int_x^\infty (t-[t])\left(\dfrac{1-3\log {t}}{t^4}\right)dt\right|\leq 2 \int_x^\infty \dfrac{\log t}{t^4}\,dt}$

$\left | \int_{x}^{\infty}(t-[t])(\frac{1-3\log {t}}{t^4})dt \right |\leq \int_{x}^{\infty}\frac{1-3\log {t}}{t^4}dt=-\frac{\log {x}}{x^3}=O(\frac{\log {x}}{x^3})$

 

[fresh_42]

$\left | \int_{x}^{\infty}(t-[t])(\frac{1-3\log {t}}{t^4})dt \right |\leq \int_{x}^{\infty}\frac{1-3\log {t}}{t^4}dt=-\frac{\log {x}}{x^3}=O(\frac{\log {x}}{x^3})$

Why? This cannot be since we have a positive number on the left and a negative on the right!
\begin{align*}
\left | \int_{x}^{\infty}(t-[t])\dfrac{1-3\log {t}}{t^4}dt \right | &\leq \int_{x}^\infty \left|t-[t]\right|\cdot \left|\dfrac{1-3\log {t}}{t^4} \right|\,dt \\
& \leq \int_{x}^\infty \left|\dfrac{1-3\log {t}}{t^4}\right|\,dt \\
&\leq \int_{x}^\infty \dfrac{1}{t^4}\,dt + 3 \int_{x}^\infty \dfrac{\log t}{t^4} \,dt\\
&\leq \dfrac{1}{3x^3} + \infty
\end{align*}
This means that we must take more care. We only have
\begin{align*}
\left | \int_{x}^{\infty}(t-[t])\dfrac{1-3\log {t}}{t^4}dt \right | &\leq \int_{x}^\infty \left|t-[t]\right|\cdot \left|\dfrac{1-3\log {t}}{t^4} \right|\,dt \\
& \leq \int_{x}^\infty \left|\dfrac{1-3\log {t}}{t^4}\right|\,dt
\end{align*}
And now? We only know that $x>1$. So why have you split the integrals at all? What about the first integral $\int_1^\infty$ and how to deal with the zero of $1-3\log t$?

 

[Math100]

Why? This cannot be since we have a positive number on the left and a negative on the right!
\begin{align*}
\left | \int_{x}^{\infty}(t-[t])\dfrac{1-3\log {t}}{t^4}dt \right | &\leq \int_{x}^\infty \left|t-[t]\right|\cdot \left|\dfrac{1-3\log {t}}{t^4} \right|\,dt \\
& \leq \int_{x}^\infty \left|\dfrac{1-3\log {t}}{t^4}\right|\,dt \\
&\leq \int_{x}^\infty \dfrac{1}{t^4}\,dt + 3 \int_{x}^\infty \dfrac{\log t}{t^4} \,dt\\
&\leq \dfrac{1}{3x^3} + \infty
\end{align*}
This means that we must take more care. We only have
\begin{align*}
\left | \int_{x}^{\infty}(t-[t])\dfrac{1-3\log {t}}{t^4}dt \right | &\leq \int_{x}^\infty \left|t-[t]\right|\cdot \left|\dfrac{1-3\log {t}}{t^4} \right|\,dt \\
& \leq \int_{x}^\infty \left|\dfrac{1-3\log {t}}{t^4}\right|\,dt
\end{align*}
And now? We only know that $x>1$. So why have you split the integrals at all? What about the first integral $\int_1^\infty$ and how to deal with the zero of $1-3\log t$?

Sorry, I made many mistakes earlier.
$\int_{1}^{\infty}(t-[t])(\frac{1-3\log {t}}{t^4})dt=\int_{1}^{\infty}\frac{1-3\log {t}}{t^4}dt=\int_{1}^{\infty}\frac{1}{t^4}dt-\int_{1}^{\infty}\frac{3\log {t}}{t^4}dt=\frac{1}{3}-\frac{1}{3}=0$
Also, the "[t]" from my textbook has no upper part when writing this, do you know the correct Latex command for this? And what does this symbolize/indicate?

 

[fresh_42]

Sorry, I made many mistakes earlier.
$\int_{1}^{\infty}(t-[t])(\frac{1-3\log {t}}{t^4})dt=\int_{1}^{\infty}\frac{1-3\log {t}}{t^4}dt=\int_{1}^{\infty}\frac{1}{t^4}dt-\int_{1}^{\infty}\frac{3\log {t}}{t^4}dt=\frac{1}{3}-\frac{1}{3}=0$
Also, the "[t]" from my textbook has no upper part when writing this, do you know the correct Latex command for this? And what does this symbolize/indicate?

And how did you get rid of $(t-[t])$ in the integrand?

You could use
\begin{align*}
\left|\int_1^\infty (t-[t])\left(\dfrac{1-3\log t}{t^4}\right)\right|\,dt &\leq \int_1^\infty |(t-[t])|\left|\dfrac{1-3\log t}{t^4}\right|\,dt\\
&\leq \int_1^\infty \left|\dfrac{1-3\log t}{t^4}\right|\,dt\\
&= \dfrac{2}{3e}\\
&= O\left(\dfrac{\log x}{x^3}\right)
\end{align*}
and explain how the integral is achieved, and you need the triangle inequality in your original formula since the absolute values have to come in from somewhere. But one step after the other. We still need an upper bound for the $\int_x^\infty$ part.
\lfoor 5.3 \rfloor = [5.3] =5\, , \,\lceil 5.3 \rceil = 6 results in $\lfloor 5.3 \rfloor = [5.3] =5\, , \,\lceil 5.3 \rceil = 6$
This means the use of $[\ldots]$ is o.k.

 

[Math100]

And how did you get rid of $(t-[t])$ in the integrand?

You could use
\begin{align*}
\left|\int_1^\infty (t-[t])\left(\dfrac{1-3\log t}{t^4}\right)\right|\,dt &\leq \int_1^\infty |(t-[t])|\left|\dfrac{1-3\log t}{t^4}\right|\,dt\\
&\leq \int_1^\infty \left|\dfrac{1-3\log t}{t^4}\right|\,dt\\
&= \dfrac{2}{3e}\\
&= O\left(\dfrac{\log x}{x^3}\right)
\end{align*}
and explain how the integral is achieved, and you need the triangle inequality in your original formula since the absolute values have to come in from somewhere. But one step after the other. We still need an upper bound for the $\int_x^\infty$ part.
\lfoor 5.3 \rfloor = [5.3] =5\, , \,\lceil 5.3 \rceil = 6 results in $\lfloor 5.3 \rfloor = [5.3] =5\, , \,\lceil 5.3 \rceil = 6$
This means the use of $[\ldots]$ is o.k.

I do not know either. I just ended up getting rid of it to evaluate the integral without it, and to see if it matches the answer. But do you know how to? And without the upper part in bracket, the floor function, how do these values work in here?

 

[fresh_42]

From the beginning!

You have as relevant equations not none, but
$$
\sum_{y<n\leq x}f(n)=\int_{y}^{x}f(t)dt+\int_{y}^{x}(t-[t])f'(t)dt+f(x)([x]-x)-f(y)([y]-y)
$$

Applied to our sum, we get
\begin{align*}
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}&=\int_{1}^{x}\dfrac{ \log {t}}{t^3}\,dt+\int_{1}^{x}(t-[t])\left(\dfrac{ \log {t}}{t^3}\right)'\,dt+\dfrac{ \log {x}}{x^3}([x]-x)-\dfrac{ \log {1}}{1^3}([1]-1)\\
&=\left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right)+\int_1^x(t-[t])\dfrac{1-3\log t}{t^4}\,dt +O\left(\dfrac{\log x}{x^3}\right)\\
&= \left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right)+\int_1^2(t-[t])\dfrac{1-3\log t}{t^4}\,dt+\underbrace{\int_2^x(t-|t|)\dfrac{1-3\log t}{t^4}\,dt}_{=O(x^{-3}\log x)}\\
&\phantom{=}+O\left(\dfrac{\log x}{x^3}\right)\\
&= \left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right) +\dfrac{\log(16)-3}{16}+O\left(\dfrac{\log x}{x^3}\right)\\
&=\underbrace{\dfrac{1}{4}+\dfrac{\log(16)-3}{16}}_{=:A}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}+O\left(\dfrac{\log x}{x^3}\right)
\end{align*}

I used WolframAlpha for the integrals. You should calculate them explicitly!

 

[Math100]

From the beginning!

You have as relevant equations not none, but
$$
\sum_{y<n\leq x}f(n)=\int_{y}^{x}f(t)dt+\int_{y}^{x}(t-[t])f'(t)dt+f(x)([x]-x)-f(y)([y]-y)
$$

Applied to our sum, we get
\begin{align*}
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}&=\int_{1}^{x}\dfrac{ \log {t}}{t^3}\,dt+\int_{1}^{x}(t-[t])\left(\dfrac{ \log {t}}{t^3}\right)'\,dt+\dfrac{ \log {x}}{x^3}([x]-x)-\dfrac{ \log {1}}{1^3}([1]-1)\\
&=\left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right)+\int_1^x(t-[t])\dfrac{1-3\log t}{t^4}\,dt +O\left(\dfrac{\log x}{x^3}\right)\\
&= \left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right)+\int_1^2(t-[t])\dfrac{1-3\log t}{t^4}\,dt+\underbrace{\int_2^x(t-|t|)\dfrac{1-3\log t}{t^4}\,dt}_{=O(x^{-3}\log x)}\\
&\phantom{=}+O\left(\dfrac{\log x}{x^3}\right)\\
&= \left(\dfrac{1}{4}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}\right) +\dfrac{\log(16)-3}{16}+O\left(\dfrac{\log x}{x^3}\right)\\
&=\underbrace{\dfrac{1}{4}+\dfrac{\log(16)-3}{16}}_{=:A}-\dfrac{\log x}{2x^2}-\dfrac{1}{4x^2}+O\left(\dfrac{\log x}{x^3}\right)
\end{align*}

I used WolframAlpha for the integrals. You should calculate them explicitly!

But how does $\frac{\log {x}}{x^3}([x]-x)=O(\frac{\log {x}}{x^3})$? And how to find $\int_{1}^{2}(t-[t])\frac{1-3\log {t}}{t^4}dt, \int_{2}^{x}(t-[t])\frac{1-3\log {t}}{t^4}dt$?

 

[fresh_42]

But how does $\frac{\log {x}}{x^3}([x]-x)=O(\frac{\log {x}}{x^3})$?

$-\dfrac{\log {x}}{x^3} \leq \dfrac{\log {x}}{x^3}([x]-x) \leq 0$
It gets as negative as $\dfrac{\log {x}}{x^3}$ gets.

The entire expression is positive. We must show that
$$
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}=A-\dfrac{ \log {x}}{2x^2}-\dfrac{1}{4x^2}+O\left(\dfrac{ \log {x}}{x^3}\right)
$$
which means
$$
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}=A-\dfrac{ \log {x}}{2x^2}-\dfrac{1}{4x^2}+C\cdot \left(\dfrac{ \log {x}}{x^3}\right)
$$
with some constant $C$. That means that $C$ swallows all coefficients in front of terms $\dfrac{\log x}{x^3}$ whether they are negative, positive, or zero. The result, in the end, will be positive, but even this is irrelevant.

And how to find $\int_{1}^{2}(t-[t])\frac{1-3\log {t}}{t^4}dt, \int_{2}^{x}(t-[t])\frac{1-3\log {t}}{t^4}dt$?

As I said, I used WolframAlpha, I haven't tried to integrate them. The split at $2$ was because the first integral $\int_1^2$ has the zero of the function, so I needed an exact value. The right integrand is always zero or negative, so
$$
0\geq (t-[t])\dfrac{1-3\log {t}}{t^4} \geq \dfrac{1-3\log {t}}{t^4}
$$
and
$$
\int_{2}^{x} 0\,dt = 0 \geq \int_{2}^{x}(t-[t])\dfrac{1-3\log {t}}{t^4}\,dt \geq \int_{2}^{x}\dfrac{1-3\log {t}}{t^4}\,dt =\dfrac{\log x}{x^3}-\dfrac{\log 2}{8}
$$
which is again $O\left(\dfrac{ \log {x}}{x^3}\right)$ and a constant that ends up in $A.$ I should have better defined $A=\dfrac{1}{4}+\dfrac{\log (16) -3}{16}-\dfrac{\log(2)}{8}=\dfrac{1}{16}+\dfrac{\log(2)}{8}.$

 

[Math100]

$-\dfrac{\log {x}}{x^3} \leq \dfrac{\log {x}}{x^3}([x]-x) \leq 0$
It gets as negative as $\dfrac{\log {x}}{x^3}$ gets.

The entire expression is positive. We must show that
$$
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}=A-\dfrac{ \log {x}}{2x^2}-\dfrac{1}{4x^2}+O\left(\dfrac{ \log {x}}{x^3}\right)
$$
which means
$$
\sum_{n\leq x}\dfrac{ \log {n}}{n^3}=A-\dfrac{ \log {x}}{2x^2}-\dfrac{1}{4x^2}+C\cdot \left(\dfrac{ \log {x}}{x^3}\right)
$$
with some constant $C$. That means that $C$ swallows all coefficients in front of terms $\dfrac{\log x}{x^3}$ whether they are negative, positive, or zero. The result, in the end, will be positive, but even this is irrelevant.
As I said, I used WolframAlpha, I haven't tried to integrate them. The split at $2$ was because the first integral $\int_1^2$ has the zero of the function, so I needed an exact value. The right integrand is always zero or negative, so
$$
0\geq (t-[t])\dfrac{1-3\log {t}}{t^4} \geq \dfrac{1-3\log {t}}{t^4}
$$
and
$$
\int_{2}^{x} 0\,dt = 0 \geq \int_{2}^{x}(t-[t])\dfrac{1-3\log {t}}{t^4}\,dt \geq \int_{2}^{x}\dfrac{1-3\log {t}}{t^4}\,dt =\dfrac{\log x}{x^3}-\dfrac{\log 2}{8}
$$
which is again $O\left(\dfrac{ \log {x}}{x^3}\right)$ and a constant that ends up in $A.$ I should have better defined $A=\dfrac{1}{4}+\dfrac{\log (16) -3}{16}-\dfrac{\log(2)}{8}=\dfrac{1}{16}+\dfrac{\log(2)}{8}.$

Also, how did you get $\frac{\log {16}-3}{16}$? Did the $-\frac{\log {2}}{8}$ get swallowed by A or C?

 

[fresh_42]

Also, how did you get $\frac{\log {16}-3}{16}$?

WA

Did the $-\frac{\log {2}}{8}$ get swallowed by A or C?

A is easier.

 

[Math100]

WA

A is easier.

WA?

 

[fresh_42]

WA?

https://www.wolframalpha.com/input?i=integral+(from+1+to+2)+((x-+entier(x))(1-3log+x)/(x^4)+)dx+=

 

[fresh_42]

WA?

It's not difficult. You already did it, too.
$$
\int_1^2 (x-[x])\cdot \dfrac{1-3\log x}{x^4}\,dt= \underbrace{\int_1^2\dfrac{1-3\log x}{x^3}\,dt}_{=3(\log(4)-1)/16}- \underbrace{\int_1^2\dfrac{1-3\log x}{x^4}\,dt}_{=\log(2)/8}
$$

 

[Math100]

It's not difficult. You already did it, too.
$$
\int_1^2 (x-[x])\cdot \dfrac{1-3\log x}{x^4}\,dt= \underbrace{\int_1^2\dfrac{1-3\log x}{x^3}\,dt}_{=3(\log(4)-1)/16}- \underbrace{\int_1^2\dfrac{1-3\log x}{x^4}\,dt}_{=\log(2)/8}
$$

$\int_{1}^{2}(t-[t])\frac{1-3\log {t}}{t^4}dt$
$=\int_{1}^{2}\frac{1-3\log {t}}{t^3}dt-\int_{1}^{2}\frac{1-3\log {t}}{t^4}dt$
$=[-\frac{3}{16}+\frac{3}{8}\log {2}]-\frac{1}{8}\log {2}$
$=-\frac{3}{16}+\log {2^{\frac{1}{4}}}=-\frac{3-\log {16}}{16}$