3

I want to show that$\displaystyle\lim_{s \to 1^+}\int_{n}^{n+1}\cfrac{1}{x^s}dx = \int_{n}^{n+1}\cfrac{1}{x}dx$.

I find the left hand side as: $\lim_{s \to 1^+}(s-1)(\cfrac{1}{n^s}-\cfrac{1}{(n+1)^s})$. However, if I am not wrong, the right hand side is equal to $\ln(n+1)-\ln(n)$.

Do I miss something? How can I show this equality?

My main goal is to understand the equality $ \lim_{s \rightarrow 1^+} \left( \zeta(s) - \frac{1}{s-1} \right) = \gamma$ that is asked at Limit of Zeta function.

Edit: Thank you all for your valuable comments and answers. We clearly have $ \lim_{s \rightarrow 1^+} \frac{1}{s-1} =\infty$. Since $\frac{1}{s-1}$ diverges, I can not seperate the limits. Now I would like to conclude that $ \lim_{s \rightarrow 1^+} \zeta(s) = \infty$. I need some hints because, unfortunately, I can not see how to get the result.

Ninja
  • 2,797
  • 1
    Can't you just switch the order of $\lim$ and $\int$? Everything is continuous here except when $x = 0$ – GFauxPas Jul 24 '17 at 12:37
  • 2
    Note that the integral on the left hand side evaluates to $\frac{1}{s-1}\left(\frac{1}{n^{s-1}} - \frac{1}{(n+1)^{s-1}}\right)$ so it's $s-1$ instead of $s$ in the exponents and $(s-1)$ is in the denominator so you get a $0\cdot \infty$ limit as $s\to 1$ that you can evaluate. – Winther Jul 24 '17 at 12:37
  • I think the equality you want involves $\lim_{s\to 1^+}$ and not $\lim_{s\to 1}$. – Masacroso Jul 24 '17 at 12:41
  • @Masacroso, true, I have edited the last line. – Ninja Jul 24 '17 at 12:45
  • 1
    Hy hint would be: show: $1/x^s$ converges to $1/x$ as $s \to 1^+$ uniformly on $[n,n+1]$. Of course this requires either $n>0$ or $n+1 < 0$. – GEdgar Jul 24 '17 at 16:23
  • @GEdgar, excuse me but I could not show the equality. – Ninja Jul 25 '17 at 06:34

3 Answers3

1

For $s>1$, an antiderivative of $x^{-s}$ is $$ \frac{x^{-s+1}}{-s+1} $$ so the integral on the left-hand side is $$ \frac{(n+1)^{1-s}}{1-s}-\frac{n^{1-s}}{1-s}= \frac{(n+1)^{1-s}-n^{1-s}}{1-s} $$ (this shows you computed the integral wrongly).

Thus you need $$ \lim_{s\to 1^+}\frac{(n+1)^{1-s}-n^{1-s}}{1-s}= \lim_{t\to 0^-}\frac{(n+1)^{t}-n^t}{t} $$ which is the derivative at $0$ of the function $$ f(t)=(n+1)^{t}-n^t $$ Since $$ f'(t)=(n+1)^{t}\ln(n+1)-n^t\ln n $$ you have $$ f'(0)=\ln(n+1)-\ln n=\int_{n}^{n+1}\frac{1}{x}\,dx $$

egreg
  • 238,574
0

Why not use Lebesgue dominated convergence theorem?

For $n \in \mathbb N$, $x \in [n, n+1]$ and $s \ge 1$ you have $$0 \le \frac{1}{x^s} \le \frac{1}{x} \text{ and } \lim\limits_{s \to 1^+} \frac{1}{x^s} = \frac{1}{x}.$$

Therefore according to Lebesgue dominated convergence theorem $$\lim\limits_{s \to 1^+} \int_n^{n+1} \frac{dx}{x^s} = \int_n^{n+1} \frac{dx}{x} = \ln(n+1)-\ln(n)$$

mathcounterexamples.net
  • 70,018
-1

The easiest way to see this doesn't involve any fancy algebraic tricks like the paths you tried. Rather, since the argument of the integral approaches 'nicely' enough, so must its argument: in particular, here, the integrand is uniformly continuous and monotonic, which are both very nice properties for taking limits. Merely verifying that the integral is well-defined at all s>1 is then enough. I don't cite a particular theorem from real analysis because there are many different ones that could apply, depending on which properties you particularly choose to use.

I do find it easiest to imagine graphically, though: if you have the 'regions' in the xy plane for each of the s>1, then taking their (nicely measurable) intersection yields the region for the right-hand-side integral.

Alex Meiburg
  • 2,503
  • 11
  • 21