$-\int_{1}^{\infty}\sum_{n=0}^{\infty}(n+x)^{-3}dx$

Calculus (Integrals, Series)
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Grigorios Kostakos
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$-\int_{1}^{\infty}\sum_{n=0}^{\infty}(n+x)^{-3}dx$

#1

Post by Grigorios Kostakos »

Evaluate:

$$\int_{1}^{+\infty}\mathop{\sum}\limits_{n=0}^{+\infty} \frac{-1}{(n+x)^3}\,{\rm{d}}x$$
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Re: $-\int_{1}^{\infty}\sum_{n=0}^{\infty}(n+x)^{-3}dx$

#2

Post by Tolaso J Kos »

Grigorios Kostakos wrote:Evaluate:

$$\int_{1}^{+\infty}\mathop{\sum}\limits_{n=0}^{+\infty} \frac{-1}{(n+x)^3}\,{\rm{d}}x$$
Hello Grigoris...

The polygamma derivatives are given by the formula:

$$\psi^{(n)}(x)=(-1)^{n+1}n! \sum_{k=0}^{\infty}\frac{1}{\left ( x+k \right )^{n+1}}$$

For $n=2$ we get that:

$$\psi^{(2)}(x)=2!\sum_{k=0}^{\infty}\frac{-1}{\left ( k+x \right )^3} \Leftrightarrow \sum_{k=0}^{\infty}\frac{-1}{\left ( k+x \right )^3}=\frac{\psi^{(2)}(x)}{2}$$

Hence the origininal question is reduced down to:

$$\begin{aligned}
\int_{1}^{\infty}\left [ -\sum_{k=0}^{\infty}\frac{1}{\left ( x+k \right )^3} \right ]\,dx &= \int_{1}^{\infty}\frac{\psi^{(2)}(x)}{2}\,dx\\
&= \frac{1}{2}\left [ \psi^{(1)}(x) \right ]_1^\infty\\
&= \frac{1}{2}\psi^{(1)}(\infty)-\frac{1}{2}\psi^{(1)}(1)\\
&=0-\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{\left ( n+1 \right )^2} \\
&= -\frac{1}{2}\cdot \frac{\pi^2}{6}=-\frac{\pi^2}{12}
\end{aligned}$$

and we are done. Do we have an alternative?
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Re: $-\int_{1}^{\infty}\sum_{n=0}^{\infty}(n+x)^{-3}dx$

#3

Post by Grigorios Kostakos »

Tolis,

I had in mind two solutions. The one you have given and another by interchanging integral and sum. But for the second solution is required to justify the interchange, which is the difficult point of these solution.
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Re: $-\int_{1}^{\infty}\sum_{n=0}^{\infty}(n+x)^{-3}dx$

#4

Post by Papapetros Vaggelis »

Hello.

Let \(\displaystyle{\left(f_{n}\right)_{n\in\mathbb{N}\cup\left\{0\right\}}}\) be the sequence of the functions

\(\displaystyle{f_{n}:\left[1,+\infty\right)\longrightarrow \mathbb{R}}\) given by \(\displaystyle{f_{n}(x)=\dfrac{-1}{\left(n+x\right)^3}}\) .

The functions \(\displaystyle{f_{n}\,,n\in\mathbb{N}\cup\left\{0\right\}}\) are continuous and infinitely many times differentiable at \(\displaystyle{\left[1,+\infty\right)}\) .

For each \(\displaystyle{x\geq 1}\) holds : \(\displaystyle{\lim_{n\to \infty}f_{n}(x)=\lim_{n\to \infty}\dfrac{-1}{\left(n+x\right)^3}=0}\), so

the sequence \(\displaystyle{\left(f_{n}\right)_{n\in\mathbb{N}\cup\left\{0\right\}}}\) converges pointwise to the function

\(\displaystyle{f:\left[1,+\infty\right)\longrightarrow \mathbb{R}\,,f(x)=0}\) .

\(\displaystyle{\left(\forall\,n\in\mathbb{N}\right)\,\left(\forall\,x\geq 1\right) : \left|f_{n}(x)-f(x)\right|=\left|f_{n}(x)\right|=\dfrac{1}{\left(n+x\right)^3}}\).

\(\displaystyle{\left(\forall\,n\in\mathbb{N}\right)\,\left(\forall\,x\geq 1\right): n+x\geq n+1\implies \left(n+x\right)^3\geq \left(n+1\right)^3\implies \dfrac{1}{\left(n+x\right)^3}\leq \dfrac{1}{\left(n+1\right)^3}}\)

and thus: \(\displaystyle{\left|f_{n}(x)\right|\leq \dfrac{1}{\left(n+1\right)^3}\,,\forall\,n\in\mathbb{N}\,,\forall\,x\geq 1}\) where :

\(\displaystyle{\sum_{n=0}^{\infty}\dfrac{1}{\left(n+1\right)^3}=\sum_{n=1}^{\infty}\dfrac{1}{n^3}\leq \sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}<\infty}\) .

Therefore, the sequence \(\displaystyle{\left(f_{n}\right)_{n\in\mathbb{N}\cup\left\{0\right\}}}\) converges uniformly to \(\displaystyle{f}\) and

for \(\displaystyle{x\geq 1}\) :

\(\displaystyle{\int_{1}^{x}\sum_{n=0}^{\infty}\dfrac{-1}{\left(n+t\right)^3}\,\mathrm{d}t=\sum_{n=0}^{\infty}\,\int_{1}^{x}\dfrac{-1}{\left(n+t\right)^3}\,\mathrm{d}t}\)

where:

\(\displaystyle{\int_{1}^{x}\dfrac{-1}{\left(n+t\right)^3}\,\mathrm{d}t=\left[\dfrac{1}{2\,\left(n+t\right)^2}\right]_{1}^{x}=\left[\dfrac{1}{2\,\left(n+x\right)^2}-\dfrac{1}{2\,\left(n+1\right)^2}\right]}\) .

Taking all things above into consideration, we get :

\(\displaystyle{\forall\,x\geq 1: \int_{1}^{x}\sum_{n=0}^{\infty}\dfrac{-1}{\left(n+t\right)^3}\,\mathrm{d}t=\sum_{n=0}^{\infty}\left[\dfrac{1}{2\,\left(n+x\right)^2}-\dfrac{1}{2\,\left(n+1\right)^2}\right]}\) .

\(\displaystyle{\bullet }\) \(\displaystyle{\sum_{n=0}^{\infty}\dfrac{1}{2\,\left(n+1\right)^2}=\dfrac{1}{2}\,\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{12}}\) .

\(\displaystyle{\bullet }\) Since the convergence is uniform, we have that :

\(\displaystyle{\lim_{x\to +\infty}\,\sum_{n=0}^{\infty}\dfrac{1}{2\,\left(n+x\right)^2}=\sum_{n=0}^{\infty}\,\lim_{x\to +\infty}\dfrac{1}{2\,\left(n+x\right)^2}=0}\) .

According to the algebra of limits,

\(\displaystyle{\begin{aligned}\lim_{x\to +\infty}\int_{1}^{x}\sum_{n=0}^{\infty}\dfrac{-1}{\left(n+t\right)^3}\,\mathrm{d}t&=\lim_{x\to +\infty}\sum_{n=0}^{\infty}\left[\dfrac{1}{2\,\left(n+x\right)^2}-\dfrac{1}{2\,\left(n+1\right)^2}\right]\\&=\lim_{x\to +\infty}\sum_{n=0}^{\infty}\dfrac{1}{\left(n+x\right)^2}-\lim_{x\to +\infty}\,\sum_{n=0}^{\infty}\dfrac{1}{2\,\left(n+1\right)^2}\\&=\lim_{x\to +\infty}\sum_{n=0}^{\infty}\dfrac{1}{\left(n+x\right)^2}-\lim_{x\to +\infty}\,\left(\dfrac{\pi^2}{12}\right)\\&=0-\dfrac{\pi^2}{12}\\&=-\dfrac{\pi^2}{12}\end{aligned}}\)
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