## § 3.3: Euler's summation formula

### Main idea behind the proof of Euler's summation formula

Important numbers in the proof: \[ 0, \quad \underbrace{[y]}_{=\,m}, \quad y, \quad \underbrace{[y] + 1}_{=\,m + 1}, \quad \underbrace{[x]}_{=\,k}, \quad x. \] Splitting the definite integral: \[ \int_y^x f(t)\,dt = \int_{y}^{[y] + 1} f(t)\,dt + \underbrace{\int_{[y] + 1}^{[y] + 2} f(t)\,dt + \dots + \int_{[x] - 1}^{[x]} f(t)\,dt}_{=\,\int_{[y] + 1}^{[x]} f(t)\, dt} + \int_{[x]}^{x} f(t)\,dt. \] Using the more convenient variables \( m \) and \( k \), we get: \[ \int_y^x f(t)\,dt = \int_m^{m + 1} f(t)\,dt + \underbrace{\int_{m + 1}^{m + 2} f(t)\,dt + \dots + \int_{k - 1}^{k} f(t)\,dt}_{=\,\int_{m + 1}^{k} f(t)\, dt} + \int_{k}^{x} f(t)\,dt. \]

### Sum of integrals in the proof Euler's summation formula

\begin{align*} \int_{m + 1}^{k} [t] f'(t) dt & = \int_{m + 1}^{m + 2} [t] f'(t) dt + \int_{m + 2}^{m + 3} [t] f'(t) dt + \dots + \int_{k - 1}^{k} [t] f'(t) dt \\ & = \begin{aligned}[t] & (m + 2) f(m + 2) - (m + 1) f(m + 1) - f(m + 2) \\ + & (m + 3) f(m + 3) - (m + 2) f(m + 2) - f(m + 3) \\ & \dots \\ + & (k) f(k) - (k - 1) f(k - 1) - f(k) \end{aligned} \\ & = kf(k) - (m + 1)f(m + 1) - \sum_{n=m + 2}^{k} f(n) \\ & = kf(k) - mf(m + 1) - f(m + 1) - \sum_{n=m + 2}^{k} f(n) \\ & = kf(k) - mf(m + 1) - \sum_{n=m + 1}^{k} f(n) \\ & = kf(k) - mf(m + 1) - \sum_{y \lt n \le x} f(n). \end{align*}

### Equation (6) in the proof of Euler's summation formula

\begin{align*} \sum_{y \lt n \le x} f(n) & = - \int_{m + 1}^k [t] f'(t) \, dt + k f(k) - m f(m + 1) \\ & = \begin{aligned}[t] & \left( - \int_y^{m + 1} [t] f'(t) \, dt - \int_{m + 1}^k [t] f'(t) \, dt - \int_k^x [t] f'(t) \, dt \right) \\ & + f(k) - m f(m + 1) + \int_y^{m + 1} [t] f'(t) \, dt + \int_k^x [t] f'(t) \, dt \end{aligned} \\ & = - \int_y^x [t] f'(t) \, dt + k f(k) - m f(m + 1) + \int_y^{m + 1} m f'(t) \, dt + \int_k^x k f'(t) \, dt \\ & = - \int_y^x [t] f'(t) \, dt + k f(k) - m f(m + 1) + \biggl( m f(m + 1) - m f(y) \biggr) + \biggl( k f(x) - k f(k) \biggr) \\ & = - \int_y^x [t] f'(t) \, dt + k f(x) - m f(y). \end{align*}

### Using integration by parts in the proof of Euler's summation formula

Integration by parts: \[ \int uv \, dt = u \int v \, dt - \int u' \left( \int v \, dt \right) \, dt. \] \[ \int_y^x t f'(t) \, dt = \left. \left( t f(t) - \int f(t) \, dt \right) \right|_y^x = x f(x) - y f(y) - \int_y^x f(t) \, dt. \] Final step of the proof: \begin{align*} \sum_{y \lt n \le x} f(n) & = -\int_y^x [t] f'(t) \, dt + k f(x) - m f(y) \\ & = \begin{aligned}[t] & -\int_y^x [t] f'(t) \, dt + [x] f(x) - [y] f(y) \\ & + \underbrace{ \left( \int_y^x t f'(t) \, dt - x f(x) + y f(y) + \int_y^x f(t) \, dt \right)}_{0 \text{ by above definite integral}} \end{aligned} \\ & = \int_y^x f(t) \, dt + \int_y^x (t - [t]) f'(t) \, dt + f(x)([x] - x) - f(y)([y] - y). \end{align*}

## § 3.3: Some elementary asymptotic formulas

### Splitting integral in the proof of Theorem 3.2

Splitting definite integral: \begin{align*} & \int_1^{\infty} f(t) \, dt = \int_1^{x} f(t) \, dt + \int_x^{\infty} f(t) \, dt \\ & \iff \int_1^{\infty} f(t) \, dt - \int_x^{\infty} f(t) \, dt = \int_1^x f(t) \, dt. \end{align*} Solving improper integral: \[ \int_x^{\infty} \frac{1}{t^2} \, dt = \lim_{b \to \infty} \int_x^b \frac{1}{t^2} dt = \lim_{b \to \infty} \frac{-1}{t} \Biggr|_x^b = \left( \lim_{b \to \infty} \frac{-1}{b} \right) + \frac{1}{x} = 0 + \frac{1}{x} = \frac{1}{x}. \]

### Euler's constant in the proof of Theorem 3.2 (a)

Definition of Euler's constant: \[ C = \lim_{n \to \infty} \left( 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} - \log n \right) = \lim_{x \to \infty} \left( \sum_{n \le x} \frac{1}{n} - \log x \right). \] We begin with \[ \sum_{n \le x} \frac{1}{n} = \log x + \underbrace{1 - \int_1^{\infty} \frac{t - [t]}{t^2} \, dt}_{\text{We will show below that this is \( C \)}} + O\left( \frac{1}{x} \right). \] Rearranging the terms, we get \[ \sum_{n \le x} \frac{1}{n} - \log x = 1 - \int_1^{\infty} \frac{t - [t]}{t^2} \, dt + O\left( \frac{1}{x} \right). \] Using the definition of \( C \), we get \begin{align*} C & = \lim_{x \to \infty} \left( \sum_{n \le x} \frac{1}{n} - \log x \right) \\ & = \lim_{x \to \infty} \left( 1 - \int_1^{\infty} \frac{t - [t]}{t^2} \, dt + O\left( \frac{1}{x} \right) \right) \\ & = 1 - \int_1^{\infty} \frac{t - [t]}{t^2} \, dt. \end{align*}

### Some integrals in the proof of Theorem 3.2 (b)

\[ \int_1^x \frac{dt}{t^s} = \frac{t^{-s + 1}}{-s + 1} \Biggr|_1^x = \frac{t^{1 - s}}{1 - s} \Biggr|_1^x = \frac{x^{1 - s}}{1 - s} - \frac{1}{1 - s}. \] \[ \int_1^x \frac{t - [t]}{t^{s + 1}} \, dt = \int_1^{\infty} \frac{t - [t]}{t^{s + 1}} \, dt - \int_x^{\infty} \frac{t - [t]}{t^{s + 1}} \, dt = \int_1^{\infty} \frac{t - [t]}{t^{s + 1}} \, dt + \underbrace{\frac{1}{s} O\left( x^{-s}\right)}_{\text{explained below}}. \] \[ 0 \le \int_x^{\infty} \frac{t - [t]}{t^{s + 1}} \, dt \le \int_x^{\infty} \frac{1}{t^{s + 1}} \, dt = \frac{-1}{st^s} \Biggr|_x^\infty = \frac{1}{sx^s} = \frac{1}{s} x^{-s}. \] \begin{align*} \sum_{n \le x} \frac{1}{n^s} & = \int_1^x \frac{dt}{t^s} - s \int_1^x \frac{t - [t]}{t^{s + 1}} + 1 - \frac{x - [x]}{x^s} \, dt \\ & = \frac{x^{1 - s}}{1 - s} - \frac{1}{1 - s} - s \int_1^{\infty} \frac{t - [t]}{t^{s + 1}} \, dt + 1 + O(x^{-s}). \end{align*}