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I have from integral transform tables that:

$$\operatorname{Mellin}(\lfloor x\rfloor) = -\dfrac{\zeta(-z)}{z},\quad \operatorname{Re}(z)<-1$$

How can this be proved?

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user14685
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2 Answers2

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With the characteristic functions $\chi_n$ of the intervals $[n,\infty)$, we can write

$$\lfloor x\rfloor = \sum_{n=1}^\infty \chi_n(x),$$

and thus

$$\int_1^\infty \frac{\lfloor x\rfloor}{x^{s+1}}\,dx = \int_1^\infty \frac{1}{x^{s+1}}\sum_{n=1}^\infty \chi_n(x)\,dx,$$

and one can interchange summation and integration (dominated convergence theorem) to obtain

$$\int_1^\infty \frac{\lfloor x\rfloor}{x^{s+1}}\,dx = \sum_{n=1}^\infty \int_1^\infty \frac{\chi_n(x)}{x^{s+1}}\,dx = \sum_{n=1}^\infty \int_n^\infty \frac{dx}{x^{s+1}} = \sum_{n=1}^\infty \left[-\frac{1}{sx^s}\right]_n^\infty = \frac{1}{s}\sum_{n=1}^\infty \frac{1}{n^s} = \frac{\zeta(s)}{s}$$

for $\operatorname{Re} s > 1$. Normalize that computation to conform to the definition of the Mellin transform you're working with.

Daniel Fischer
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  • How would you define the characteristic function here? Is it 1 on the domain [n,inf)... Then \Xi_n(4.1)=...? – user14685 Apr 13 '14 at 16:19
  • Yes, $$\chi_n(x) = \begin{cases} 1 &, x \geqslant n\ 0 &, x < n, \end{cases}$$ so $\chi_n(4.1) = 0$ for $n > 4$, and $\chi_n(4.1) = 1$ for $n\in {1,2,3,4}$. – Daniel Fischer Apr 13 '14 at 16:21
2

$$ \int_{0}^{\infty} \lfloor x\rfloor x^{z-1} \ dx = \sum_{k=0}^{\infty} \int_{k}^{k+1} k x^{z-1} \ dx = \sum_{k=1}^{\infty} k \int_{k}^{k+1} x^{z-1} \ dx $$

$$ = \sum_{k=1}^{\infty} \frac{k}{z} \Big((k+1)^{z} - k^{z} \Big) = \frac{1}{z}\Big( (2^{z} - 1 ) + 2 (3^{z}-2^{z} ) + 3 (4^{z}-3^{z} ) + \ldots\Big)$$

$$ = \frac{1}{z} \Big( -1 -2^{z} - 3^{z} - \ldots) = - \frac{1}{z} \Big(1+2^{z}+3^{z} + \ldots \Big) = - \frac{\zeta(-z)}{z}$$

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