This post mainly discusses several common integration methods related to asymptotic method and the relationship between total variation and integration.

Asymptotic Expansions [1]

The principle is illustrated most easily by the integral: \[ \int_x^{\infty} e^{-t} t^{-n-r-1} d t \quad(1) \] and if \(n+r=x\) the next term is numerically equal to the last kept. But if we take the logarithmic derivative of the integrand we get \[ \frac{d}{d t}\{-t-(n+r+1) \ln t\}=-1-\frac{n+r+1}{t}\quad(2) \] and the two terms are nearly equal when \(t=x\). Hence the integral is nearly \[ e^{-x} x^{-n-r-1} \int_0^{\infty} e^{-2 u} d u=\frac{1}{2} e^{-x} x^{-n-r-1}\quad(3) \] This step may be incorrect, please see my notes below for details

and the remainder term is nearly \[ (-)^{r+1} \frac{1}{2} n(n+1) \ldots(n+r) e^{-x} x^{-n-r-1}\quad(4) \] which is half the next term in the expansion. Thus a very substantial improvement will be made if the asymptotic series is computed up to the smallest term, and half the next term is added. Greater accuracy still is obtainable by expanding $\{t+(n+r+1) \ln t\}$ to higher powers of $(t-x) / x$.


take logarithmic derivative of the integrand in equation (1) \[ \frac{d}{d t} \ln \left(e^{-t} t^{-n-r-1}\right)=\frac{d}{d t}[-t-(n+r+1) \ln t] \quad (1.1) \]

Appendix: logarithmic derivative [3]

we provide logarithmic derivative here \[ \frac{d \ln f}{d x}=\frac{f^{\prime}}{f} \]


then equation (1.1) becomes \[ -1-\frac{n+r+1}{t} \] and \(t=x\) (refer to " the two terms are nearly equal when \(t=x\)") \[ \frac{f^{\prime}}{f}=-1-\frac{n+r+1}{x} \]

Also since the next term is equal to the last kept as \(n+t=x\), then \[ \frac{f^{\prime}}{f}=-2-\frac{1}{x} \sim-2 \text { as } x \rightarrow \infty \] here \(f=e^{c_1} e^{-2 u}\), put \(u=x\) (here we treat it \(x\) as a constant). \[ \begin{aligned} &f=e^{-2 x} e^{c_1}=e^{-x} x^{-n-r-1} \\ &\Rightarrow e^{c_1}=e^x x^{-n-r-1} \end{aligned} \] we arrive at the following equation \[ e^x x^{-n-r-1} \int_x^{\infty} e^{-2 u} d u \] here is the correct form of this equation

it goes as follow \[ \begin{aligned} \Rightarrow \quad & e^x x^{-n-r-1}\left[0-\left(-\frac{1}{2} e^{-2 x}\right)\right] \\ & =e^x x^{-n-r-1} \frac{1}{2} e^{-2 x} \\ & =\frac{1}{2} e^{-x} x^{-n-r-1} \end{aligned} \]

convergence factor [4]

To illustrate, consider the following example \[ \int_{-\infty}^{\infty} \frac{\sin x}{x} d x \] we define a new integral, to represent the part of integrand, denoted by: \[ \int_0^{\infty} e^{x t} d t=\frac{1}{x} \] Therefore \[ \begin{aligned} & \int_{-\infty}^{\infty} \frac{\sin x}{x} d x=2 \int_0^{\infty} \frac{\sin x}{x} d x \\ & =2 \int_0^{\infty} \int_0^{\infty} e^{-x t} \sin x d x d t \\ & =2 \int_0^{\infty} \frac{1}{1+t^2} d t \end{aligned} \] Note: Euler's formula may facilitate the computation involving this integrand \(e^{-x t} \sin x\), it follow that \[ e^{-x t} \sin x \rightarrow e^{-x t} \frac{1}{2 i}\left(e^{i x}-e^{-i x}\right) \]



Another case we chiefly need is \[ \quad I=\int_{-A}^B e^{-1 / 2 b^2 z^2} f(z) d z \quad (5) \]

where \(A, B\) are positive and independent of \(b\), and \(f(z)\) still has an expansion given by near \(z=0\). We put \(z^2=\zeta\), \[ I=\frac{1}{2} \int_0^{A^2} e^{-1 / 2 b^2 \zeta} f\left(-\zeta^{1 / 2}\right) \zeta^{-1 / 2} d \zeta+\frac{1}{2} \int_0^{B^2} e^{-1 / 2 b^2 \zeta} f\left(\zeta^{1 / 2}\right) \zeta^{-1 / 2} d \zeta \quad (6) \]

The odd powers cancel and \[ I \sim \sqrt{2 \pi}\left(\frac{a_0}{b}+\frac{a_2}{b^3}+1\cdot 3 \frac{a_4}{b^5}+\ldots+1\cdot 3 \cdots(2 n-1) \frac{a_{2 n}}{b^{2 n+1}}\right)\quad (7) \]


Explanation for equation (6)

Let \(z^2=\zeta\) Then the upper limit of integral becomes positive while the lower limit remains at 0. The new integral \((\zeta)\) bounds encompass two cases: \(z>0\) and \(z<0\). Therefore, even though the integrand correspond to different sign of \(z\left( \pm \sqrt{\zeta}\right)\), the original integral \((z)\) is equal to the sum of half of each new integral \(\left(z \rightarrow \pm \sqrt{\zeta}\right)\).

Note for equation (7)

Rewrite equation (6) into two parts: \[ \begin{aligned} & \frac{1}{2}\int_0^{A^\prime} e^{-\frac{1}{2} b^2 \zeta} f\left(-\zeta^{\frac{1}{2}}\right) \zeta^{-\frac{1}{2}} d \zeta \\ =&\frac{1}{2}\int_0^{A^{\prime}} e^{-b^2 \zeta} \zeta^{-\frac{1}{2}}\left(a_0+a_1 \zeta^{\frac{1}{2}}+a_2 \zeta^{\frac{2}{2}}+\cdots+a_{n-1} \zeta^{\frac{n-1}{2}}\right) d \zeta\\ =&\frac{1}{2}\int_0^{A^{\prime}}e^{-b^2 \zeta}\left(a_0 \zeta^{-\frac{1}{2}}+a_1+a_2 \zeta^{\frac{1}{2}}+a_3 \zeta^{\frac{2}{2}}+\cdots+a_{n-1} \zeta^{\frac{n-2}{2}}\right) d \zeta \end{aligned}\quad (8) \]

Similarly: \[ \small\begin{aligned} &\frac{1}{2} \int_0^{B^\prime} e^{-\frac{1}{2} b^2 \zeta} f\left(-\zeta^{\frac{1}{2}}\right) \zeta^{-\frac{1}{2}} d \zeta \\ &\begin{aligned} =\frac{1}{2}\int_0^{B^\prime} e^{-\frac{1}{2} b^2 \zeta} \zeta^{-\frac{1}{2}}\left[a_0+a_1\left(-\zeta^{\frac{1}{2}}\right)+\right.& a_2\left(-\zeta^{\frac{1}{2}}\right)^2+a_3\left(-\zeta^{\frac{1}{2}}\right)^3+\cdots \\ & \left.+a_{n-1}\left(-\zeta^{\frac{1}{2}}\right)^{n-1}\right] d \zeta \\ \end{aligned}\\ & =\frac{1}{2}\int_0^{B^\prime} e^{-\frac{1}{2} b^2 \zeta}\left\{a_0 \zeta^{-\frac{1}{2}}-a_1+a_2 \zeta^{\frac{1}{2}}-a_3\left(\zeta^{\frac{1}{2}}\right)^2+a_{n-1}(-1)^{n-1}\left(\zeta^{\frac{1}{2}}\right)^{n-2}\right\}d \zeta \\ \end{aligned}\quad (9) \]

combine equation (8) and (9) then we have(as \(A^{\prime}=B^{\prime}\)) \[ \begin{aligned} & \int_0^{A^{\prime}} e^{-\frac{1}{2} b^2 \zeta}\left\{a_0 \zeta^{-\frac{1}{2}}+a_2 \zeta^{\frac{1}{2}}+a_4\zeta^{\frac{3}{2}}+\cdots+a_{2 n-2}\zeta^{\frac{2 n-2-1}{2}}\right\} d \zeta \\ =& \int_0^{\infty} e^{-\frac{1}{2} b^2 \zeta}\left\{a_0 \zeta^{-\frac{1}{2}}+a_2 \zeta^{\frac{1}{2}}+a_4 \zeta^{\frac{3}{2}}+\cdots+a_{2 n-2}\zeta^{\frac{2 n-3}{2}}\right\} d \zeta \end{aligned}\quad(10) \]

we start from the first term of the power series \(a_0\) \[ \int_0^{\infty} e^{-\frac{1}{2} b^2 \zeta} a_0 \zeta^{-\frac{1}{2}} d \zeta \quad(11) \] where let \(\sqrt{\zeta}=t\rightarrow d \zeta=2 t d t\), it follows that \[ \begin{aligned} & \int_0^{\infty} e^{-\frac{1}{2} b^2 t^2} a_0 \frac{1}{t} 2 t d t \\ = & 2 a_0 \int_0^{\infty} e^{-\frac{1}{2} b^2 t^2} d t \rightarrow \text { Gauss error function } \\ = & 2 a_0 \frac{\sqrt{2}}{b} \int_0^{\infty} e^{-\left(\frac{b}{\sqrt{2}} t\right)^2} d\left(\frac{b}{\sqrt{2}} t\right) \\ = & \frac{a_0}{b} 2 \sqrt{2} \frac{\sqrt{\pi}}{2} \\ = & \frac{a_0}{b} \sqrt{2 \pi} \end{aligned}\quad(12) \] equation (10) the \(r\text{th}\) (corresponding to \(n\)) term of this equation may be written as follows \[ \begin{aligned} & \int_0^{\infty} e^{-\frac{1}{2} b^2 \zeta} a_{2 r-2} \zeta^{\frac{2 r-3}{2}} d \zeta \quad(r=2,3,4 \ldots, n) \\ = & \int_0^{\infty}\left(\frac{e^{-\frac{1}{2} b^2} \zeta}{-\frac{1}{2} b^2}\right) a_{2 r-2} \zeta^{\frac{2 r-3}{2}} d \zeta\\ = &a_{x-2}\left\{\left.\frac{e^{-\frac{1}{2} b^2\zeta} \zeta^{\frac{2 r-3}{2}}}{-\frac{1}{2} b^2}\right|_0 ^{\infty}+\frac{2 r-3}{b^2} \int_0^{\infty} e^{-\frac{1}{2} b^2 \zeta} \zeta^{\frac{2 r-5}{2}} d \zeta\right\} \\ = &a_{2 r-2} \frac{2 r-3}{b^2} \int_0^{\infty} e^{-\frac{1}{2} b^2 \zeta} \zeta^{\frac{2 r-5}{2}} d \zeta \\ = &a_{2 r-2} \frac{(2 r-3)(2 r-5) \cdots 5 \times 3 \times 1}{b^{2(r-1)}} \int_0^{\infty} e^{-\frac{1}{2} b^2 \zeta} \zeta^{-\frac{1}{2}} d \zeta \\ = &a_{2 r-2} \frac{(2 r-3)(2 r-5) \cdots 3 \times 1}{b^{2 r-2}} \frac{\sqrt{2 \pi}}{b} \\ =&a_{2 r-2} \sqrt{2 \pi} \frac{(2 r-3)(2 r-5) \cdots 3 \times 1}{b^{2 r-1}} \end{aligned}\quad(13) \] Also we have \[ \int_0^{\infty} e^{-\frac{1}{2} b^2 \zeta^2} \zeta^{-\frac{1}{2}} d \zeta=\frac{a_0}{b} \sqrt{2 \pi}\quad(14) \] as \(r=1\). Therefore we have finally \[ I \sim \sqrt{2 \pi}\left(\frac{a_0}{b}+\frac{a_1}{b^3}+1 \cdot 3 \frac{a_4}{b^5}+\cdots+1 \cdot 3 \cdots(2 r-3) \frac{a_{2 r-2}}{b^{2 r-1}}\right)\quad(15) \] end

The second mean-value theorem for integrals

We give first a simple consequence of Abel's lemma for integrals. Let \(f(x)\) be integrable in \((a, b)\) and \(\phi(x)\) have bounded variation. Let \(P(x), N(x)\) be the positive and negative variations of \(\phi(x)\) in \((a, x)\) and let the greater of \(P(b), N(b)\) be \(\omega\). Then \(\omega-P(x), \omega-N(x)\) satisfy the conditions imposed on \(v(x)\) in Abel's lemma, and if \(h, H\) are the lower and upper bounds of \(\int_a^x f(t) d t\) for \(a \leqslant x \leqslant b\),

\[ \begin{gathered} \omega h \leqslant \int_a^b\{\omega-P(x)\} f(x) d x \leqslant \omega H, \quad (B1)\\ \omega h \leqslant \int_a^b\{\omega-N(x)\} f(x) d x \leqslant \omega H \quad (B2) \end{gathered} \]

By subtraction, \[ \left|\int_a^b f(x)\{\phi(x)-\phi(a)\} d x\right| \leqslant \omega(H-h) \quad(B3) \]

Hence \(\int_a^b f(x) \phi(x) d x\) can be replaced by \(\phi(a) \int_a^b f(x) d x\) within a known uncertainty. This result is important especially in the theory of Fourier series and integrals.

This part is related to measure theory in real analysis


\(P(x)\) and \(N(x)\) are positive and negative variations respectively then we have \[ \begin{aligned} & P(x)=\int_a^x|\phi^{\prime}(x)| d x \text { where } \phi^{\prime}(x)>0 \\ & N(x)=\int_a^x\left|\phi^{\prime}(x)\right| d x \text { where } \phi^{\prime}(x)<0 \end{aligned} \] also from definition of \(\omega\), which is given \[ \omega=\max \{P(b), N(b)\} \] then it may indicates \(\omega-P(x)\) and \(\omega-N(x)\) are both non-negative, bounded and non-increasing.

Note easy to know an increase in \(x\) (until \(b\)) leads to increase in \[ \int_a^x|\phi^{\prime} (x) | d x \]

Therefore we have Equation \((B1) (B2)\) above. Next, follow these steps: Equation \((B1)\) multiplied by \(-1\) and plus equation \((B2)\) it follows that \[ -\omega(H-h)=\omega(h-H) \leq \int_a^b[P(x)-N(x)] f(x) d x \leqslant 2 \omega(H-h) \]

Therefore follow from \(-M \leq y \leq M \Rightarrow|y| \leq M\) it can be obtained \[ \left|\int_a^b[p(x)-N(x)] f(x) d x\right| \leqslant w(H-h) \] then, let's consider the \(P(x)-N(x)\). for any set of partition \(P_k\) \[ P_k=\left\{p=\left\{x_0, \ldots, x_{n_p}\right\} \mid p \text { is a partition of }\left[a, x\right]\right\} \] based on the definition of positive and negative variation, put \[ \Delta \phi_i=\phi\left(x_i\right)-\phi\left(x_{i-1}\right) \] then when \(\Delta \phi_i>0\) the \(i\text{th}\) term for \(TV^{+}(\phi)\) and \(TV^{-}(\phi)\) are \(\Delta \phi_i, 0\) respectively. Conversely when \(\Delta \phi_i<0\) then the two terms are \(0,-\Delta \phi_i\) respectively, hence for arbitrary \(n\) in the interval \([a, x]\) \[ \begin{aligned} & N(x)=\sup \left\{0+\left(-\Delta \phi_2\right)+0+\left(-\Delta \phi_4\right)+\cdots+\left(-\Delta \phi_{n-1}\right)+0\right\} \\ & P(x)=\sup\left\{\Delta \phi_1+0+\Delta \phi_3+\cdots+0+\Delta \phi_n\right\} \\ &\begin{aligned}P(x)-N(x)&=\sup\left\{\Delta \phi_1+\Delta \phi_2+\cdots+\Delta \phi_{n-1}+\Delta \phi_n\right\}\\ &\begin{aligned}=\sup&\left\{\phi\left(x_1\right)-\phi\left(x_0\right)+\phi\left(x_2\right)-\phi\left(x_1\right)+\cdots+\right.\\ &\phi\left(x_{n-1}\right)-\phi\left(x_{n-2}\right)+\phi\left(x_n\right)-\left.\phi\left(x_{n-1}\right)\right\} \end{aligned}\\ & =\sup\left\{\phi\left(x_n\right)-\phi\left(x_0\right)\right\} \end{aligned}\\ \end{aligned} \] The interval is \([a, x]\), then we have \(\phi\left(x_n\right)=\phi(x) , \phi\left(x_0\right)=\phi(a)\) and \[ P(x)-N(x)=\sup\{\phi(x)-\phi(a)\}=\phi(x)-\phi(a) \] Note that here, we put \(\Delta \phi_1>0\) and it's the same to the case \(\Delta \phi_1<0\) finially we arrived at \[ \begin{aligned} \left|\int_a^b\left[P(x)-N(x)\right] f(x) d x\right| & =\left|\int_a^b[\phi(x)-\phi(a)] f(x) d x\right| \\ & \leqslant \omega(H-h) \end{aligned} \] Q.E.D


Appendix: Bouned Variation [2, section 5.2]

\(f(x)\) 是定义在 \([a, b]\) 上的实值函数, 作分划 \(\Delta\) : \(a=x_0<x_1<\cdots<x_n=b\) 以及相应的和 \[ v_{\Delta}=\sum_{i=1}^n\left|f\left(x_i\right)-f\left(x_{i-1}\right)\right| \quad (A1) \]

称之为 \(f(x)\)\([a, b]\) 上的变差;作 \[ \bigvee_a^b(f)=\sup \left\{v_{\Delta}: \Delta \text { 为 }[a, b] \text { 的任一分划 }\right\} \quad (A2) \]

并称它为 \(f(x)\)\([a, b]\) 上的全变差(total variation) . 若 \[ \bigvee_a^b(f)<+\infty \quad (A3) \]

则称 \(f(x)\)\([a, b]\) 上的有界变差函数 (即全体变差形成有界数集),其全体记为 BV \(([a, b])\).

一个实数集合 \(A\) ,若有一个实数 \(M\) ,使得 \(A\) 中任何数都不超过 \(M\) ,那么就称 \(M\)\(A\) 的一个上界。在所有那些上界中如果有一个最小的上界,就称为 \(A\) 的上确界。 即设有一实数集 \(A \subset R\) ,实数集 \(A\) 的上确界 \(\sup A\) 被定义为如下的数: \[ \forall a \in A \Rightarrow a \leq \sup A \quad(\text{即}\sup A\text{是}A\text{的上界}) \] Source From: Baidu Baike

total variation of a differentiable function of one variable[5]

Total variation may be expressed in two different ways, case 1 is the equation (A1) given above, case 2 is as follows

The total variation of a differentiable function \(f\), defined on an interval \([a, b] \subset \mathbb{R}\), has the following expression if \(f^{\prime}\) is Riemann integrable \[ V_a^b(f)=\int_a^b\left|f^{\prime}(x)\right| \mathrm{d} x \]

If \(f\) is differentiable and monotonic, then the above simplifies to \[ V_a^b(f)=|f(a)-f(b)| \] Negative and positive variations[6]

Let \(I \subset \mathbb{R}\) be an interval and \(\Pi\) be as in Definition 1. The negative and positive variations of \(f: I \rightarrow \mathbb{R}\) are then defined as \[ \begin{gathered} T V^{+}(f):=\sup \left\{\sum_{i=1}^N \max \left\{\left(f\left(a_{i+1}\right)-f\left(a_i\right)\right), 0\right\}:\left(a_1, \ldots, a_{N+1}\right) \in \Pi\right\} \\ T V^{-}(f):=\sup \left\{\sum_{i=1}^N \max \left\{-\left(f\left(a_{i+1}\right)-f\left(a_i\right)\right), 0\right\}:\left(a_1, \ldots, a_{N+1}\right) \in \Pi\right\} . \end{gathered} \] where \(a_1<a_2<\ldots<a_{N+1} \in I\)