User:Harry Princeton/Chain Integrals and Higher-Order Integrals

This is my work on Chain Integration and Integration by Parts (higher-order integrals). Also included is a differential analysis on cycloids.

Chain Integration Formula and Examples

See Fresnel integral.

We could approximate the tangent and secant integrals

$T(x):=\int _{0}^{x}\tan {\big (}t^{2}{\big )}~{\text{d}}t$ and ${\text{Sc}}(x):=\int _{0}^{x}\sec {\big (}t^{2}{\big )}~{\text{d}}t$

by using the Cauchy Principal Value and integration by parts:

$T(x)={\frac {\ln |\sec x^{2}|}{2x}}+\int _{0}^{x}{\frac {\ln |\sec t^{2}|~{\text{d}}t}{2t^{2}}}$ and ${\text{Sc}}(x)={\frac {\ln |\tan x^{2}+\sec x^{2}|}{2x}}+\int _{0}^{x}{\frac {\ln |\tan t^{2}+\sec t^{2}|~{\text{d}}t}{2t^{2}}}$

with the square Chain Integration formula $\int _{0}^{x}f{\big (}t^{2}{\big )}~{\text{d}}t={\frac {F{\big (}x^{2}{\big )}}{2x}}+{\frac {\int _{0}^{x}F{\big (}t^{2}{\big )}~{\text{d}}t}{2x^{2}}}$ where $F(x)=\int _{0}^{x}f(t)~{\text{d}}t$.

General Chain Integration Formula For Strictly Monotone Functions $g~$

$\int _{0}^{x}f{\big (}g(t){\big )}~{\text{d}}t={\frac {F{\big (}g(x){\big )}}{g'(x)}}+\int _{0}^{x}{\frac {F{\big (}g(t){\big )}g''(t)~{\text{d}}t}{g'(t)^{2}}}$

$\Longleftrightarrow {\frac {\text{d}}{{\text{d}}x}}\left(\int _{0}^{x}f{\big (}g(t){\big )}~{\text{d}}t\right)={\frac {\text{d}}{{\text{d}}x}}\left({\frac {F{\big (}g(x){\big )}}{g'(x)}}+\int _{0}^{x}{\frac {F{\big (}g(t){\big )}g''(t)~{\text{d}}t}{g'(t)^{2}}}\right)={\frac {F'{\big (}g(x){\big )}g'(x)^{2}-F{\big (}g(x){\big )}g''(x)}{g'(x)^{2}}}+{\frac {F{\big (}g(x){\big )}g''(x)}{g'(x)^{2}}}=F'{\big (}g(x){\big )}=f{\big (}g(x){\big )}$as expected, using the Chain Rule, antiderivative, quotient rule, and cancellation.

Higher-Order Integrals:

$\iint _{0}^{x}f{\big (}g(t){\big )}~{\text{d}}t^{2}=x\left({\frac {F\left(g\left(x\right)\right)}{g'\left(x\right)}}+\int _{0}^{x}{\frac {F\left(g\left(t\right)\right)g''\left(t\right)dt}{g'\left(t\right)^{2}}}\right)-\left({\frac {xF\left(g\left(x\right)\right)}{g'\left(x\right)}}+\int _{0}^{x}{\frac {{\big (}tF\left(g\left(t\right)\right)g''\left(t\right)-F\left(g\left(t\right)\right)g'\left(t\right){\big )}dt}{g'\left(t\right)^{2}}}\right)$

${\underset {n~{\text{times}}}{\int \dots \int _{0}^{x}}}f{\big (}g(t){\big )}~{\text{d}}t^{n}={\frac {1}{(n-1)!}}\sum _{i=1}^{n}\left[\left(-1\right)^{i-1}x^{n-i}{\binom {n-1}{i}}\left({\frac {x^{i}F\left(g\left(x\right)\right)}{g'\left(x\right)}}+\int _{0}^{x}{\frac {{\big (}t^{i}F\left(g\left(t\right)\right)g''\left(t\right)-it^{i-1}F\left(g\left(t\right)\right)g'\left(t\right){\big )}dt}{g'\left(t\right)^{2}}}\right)\right]$

These can be proved using induction by integration by parts.

In fact, integrals through CPV are defined when subtracting balanced pole functions - pure powers of $1/x$ - from original poles, results in functions with remaining possible singularities of size $O(n^{-\alpha })$ with $\alpha <1$ strict).

Notice that both $\tan x^{2}$ and $\sec x^{2}$ have simple nonzero isolated poles, limiting to scalar multiples of $[\mp 1/x]_{x=0}$. Then $\int _{1}^{x}\mp {\frac {{\text{d}}t}{t}}=\mp \ln x$ and $\int _{0}^{x}\mp \ln t~{\text{d}}t=\mp x(\ln x-1)$with $\lim _{x\rightarrow 0}\mp {\big [}x(\ln x-1){\big ]}=0$.

Therefore, ${\frac {\int _{0}^{x}F{\big (}t^{2}{\big )}~{\text{d}}t}{2x^{2}}}$ is bounded for $F(x)=\ln |\sec x|=\int _{0}^{x}\tan t~{\text{d}}t$ and $F(x)=\ln |\tan x+\sec x|=\int _{0}^{x}\sec t~{\text{d}}t$; as we have the related bounded integral$\int _{0}^{1}\mp \ln x~{\text{d}}x=\mp [x(\ln x-1)]_{0}^{1}=\pm 1$ and we can do scalar multiple comparisons. So by CPV, the above integrals are defined except at isolated poles. Graphs of these integrals for $x\geq 0$ are found below:

Graphs of $T(x)$ and ${\text{Sc}}(x)$

Integral of Tan x^2	Integral of Sec x^2
[0,15] × [-5, 5]	[0,15] × [-5, 5]

Finally, we have, approximately:

$\liminf _{x\rightarrow \infty }\int _{0}^{x}\tan {\big (}t^{2}{\big )}~{\text{d}}t=\lim _{x\rightarrow \infty }\int _{0}^{x}{\frac {\ln |\sec t^{2}|~{\text{d}}t}{2t^{2}}}\approx 0.535914345467$

${\underset {x\rightarrow \infty }{\text{lim inflection}}}\int _{0}^{x}\sec {\big (}t^{2}{\big )}~{\text{d}}t=\lim _{x\rightarrow \infty }\int _{0}^{x}{\frac {\ln |\tan t^{2}+\sec t^{2}|~{\text{d}}t}{2t^{2}}}\approx 0.836834795146$

compared to ${\sqrt {\pi /8}}\approx 0.62665706866$ for both $S(x)$ and $C(x)$.

Modulated Integrals

We could also approximate the cotangent and cosecant integrals

${\text{Ct}}(x):=\int _{0}^{x}\cot {\big (}t^{2}{\big )}~{\text{d}}t$ and ${\text{Cc}}(x):=\int _{0}^{x}\csc {\big (}t^{2}{\big )}~{\text{d}}t$

by using an integration by parts, but need to isolate a pole of order 2 at $x=0$ for each function. We do so by subtracting $1/x^{2}$ from each function to yield bounded functions $f(x)$ at $x=0$ (in fact, with $\lim _{x\rightarrow 0}f(x)=0$ for both functions!), applying the same treatment $\int _{0}^{x}f{\big (}t^{2}{\big )}~{\text{d}}t={\frac {F{\big (}x^{2}{\big )}}{2x}}+{\frac {\int _{0}^{x}F{\big (}t^{2}{\big )}~{\text{d}}t}{2x^{2}}}$ as previously (it works similarly at all other poles), and then adding the antiderivative $-1/x$ of $1/x^{2}$ back.

Since both integrals have right limit $-\infty$ as $x$ approaches $0$, we instead add constants so that critical points/inflection points of the cotangent/cosecant integrals, respectively, approach $y=0$ as $x\rightarrow \infty$ (canonicalization). We thus have:

${\text{Ct}}(x):\approx 1.29414-\left({\frac {\ln 2-\ln \left(\left|\cot x^{2}+\csc x^{2}\right|\right)-2\ln \left(\left|x\right|\right)}{2x^{2}}}+{\frac {1}{x}}+\int _{0}^{x}{\frac {\ln 2-\ln \left(\left|\cot t^{2}+\csc t^{2}\right|\right)-2\ln \left(\left|t\right|\right)}{2t^{2}}}dt\right)$

${\text{Cc}}(x):\approx {\frac {\ln 2-\ln \left(\left|\cot x^{2}+\csc x^{2}\right|\right)-2\ln \left(\left|x\right|\right)}{2x}}-{\frac {1}{x}}+\int _{0}^{x}{\frac {\ln 2-\ln \left(\left|\cot t^{2}+\csc t^{2}\right|\right)-2\ln \left(\left|t\right|\right)}{2t^{2}}}dt+0.5360919$

With graphs seen below:

Graphs of ${\text{Ct}}(x)$ and ${\text{Cc}}(x)$

Integral of Cot x^2	Integral of csc x^2
[0,15] × [-5, 5]	[0,15] × [-5, 5]

Higher Order (Stacked) Integrals

Main article: Cauchy formula for repeated integration

A general formula for the second-order real integral (second antiderivative) is

$\iint _{0}^{x}f(t)~{\text{d}}t^{2}=x\int _{0}^{x}f(t)~{\text{d}}t-\int _{0}^{x}tf(t)~{\text{d}}t$

Proof ${\frac {\text{d}}{{\text{d}}x}}\left(x\int _{0}^{x}f(t)~{\text{d}}t-\int _{0}^{x}tf(t)~{\text{d}}t\right)=xf(x)+\int _{0}^{x}f(t)~{\text{d}}t-xf(x)=\int _{0}^{x}f(t)~{\text{d}}t;$by Product Rule and Fundamental Theorem of Calculus I.

A general formula for the third-order real integral (third antiderivative) is

$\iiint _{0}^{x}f(t)~{\text{d}}t^{3}={\frac {1}{2}}\left(x^{2}\int _{0}^{x}f(t)~{\text{d}}t-2x\int _{0}^{x}tf(t)~{\text{d}}t+\int _{0}^{x}t^{2}f(t)~{\text{d}}t\right)$

Proof ${\frac {\text{d}}{{\text{d}}x}}\left({\frac {1}{2}}\left(x^{2}\int _{0}^{x}f(t)~{\text{d}}t-2x\int _{0}^{x}tf(t)~{\text{d}}t+\int _{0}^{x}t^{2}f(t)~{\text{d}}t\right)\right)={\frac {1}{2}}\left(x^{2}f(x)+2x\int _{0}^{x}f(t)~{\text{d}}t-2x^{2}f(x)-2\int _{0}^{x}tf(t)~{\text{d}}t+x^{2}f(x)\right)={\frac {1}{2}}\left(2x\int _{0}^{x}f(t)~{\text{d}}t-2\int _{0}^{x}tf(t)~{\text{d}}t\right)=\iint _{0}^{x}f(t)~{\text{d}}t^{2};$by Product Rule and Fundamental Theorem of Calculus I.

A general formula for the $n$th-order real integral ($n$th antiderivative) is

${\underset {n~{\text{times}}}{\int \cdots \int _{0}^{x}}}f(t)~{\text{d}}t^{n}={\frac {1}{(n-1)!}}\sum _{i=1}^{n}(-1)^{i}{\binom {n-1}{i}}\left[x^{n-1-i}\int _{0}^{x}t^{i}f(t)~{\text{d}}t\right]$

Proof Assume by induction, that

${\underset {n-1~{\text{times}}}{\int \cdots \int _{0}^{x}}}f(t)~{\text{d}}t^{n-1}={\frac {1}{(n-2)!}}\sum _{i=1}^{n-1}(-1)^{i}{\binom {n-2}{i}}\left[x^{n-2-i}\int _{0}^{x}t^{i}f(t)~{\text{d}}t\right]$

Then,

${\frac {\text{d}}{{\text{d}}x}}\left({\frac {1}{(n-1)!}}\sum _{i=1}^{n}(-1)^{i}{\binom {n-1}{i}}\left[x^{n-i-1}\int _{0}^{x}t^{i}f(t)~{\text{d}}t\right]\right)={\frac {1}{(n-1)!}}\sum _{i=1}^{n}(-1)^{i}{\binom {n-1}{i}}\left[x^{n-i-1}x^{i}f(x)+(n-i-1)x^{n-i-2}\int _{0}^{x}t^{i}f(t)~{\text{d}}t\right]$

$={\frac {1}{(n-1)!}}\left[\sum _{i=1}^{n}(-1)^{i}{\binom {n-1}{i}}x^{n-1}f(x)+\sum _{i=1}^{n}(-1)^{n}{\binom {n-1}{i}}\int _{0}^{x}t^{i}f(t)~{\text{d}}t\right]={\frac {1}{(n-1)!}}\left[x^{n-1}f(x){\big (}1+(-1){\big )}^{n-1}+\sum _{i=1}^{n}(-1)^{n}{\frac {(n-1)!}{i!(n-i-1)!}}(n-i-1)x^{n-i-2}\int _{0}^{x}t^{i}f(t)~{\text{d}}t\right]$

$={\frac {1}{(n-1)!}}\left[x^{n-1}f(x)\cdot 0^{n-1}+\sum _{i=1}^{n}(-1)^{n}{\frac {(n-1)!}{i!(n-i-2)!}}x^{n-i-2}\int _{0}^{x}t^{i}f(t)~{\text{d}}t\right]={\frac {1}{(n-1)!}}\left[(n-1)\sum _{i=1}^{n}(-1)^{n}{\frac {(n-2)!}{i!(n-i-2)!}}x^{n-i-2}\int _{0}^{x}t^{i}f(t)~{\text{d}}t\right]$

$={\frac {1}{(n-2)!}}\sum _{i=1}^{n-1}(-1)^{i}{\binom {n-2}{i}}\left[x^{n-2-i}\int _{0}^{x}t^{i}f(t)~{\text{d}}t\right]={\underset {n-1~{\text{times}}}{\int \cdots \int _{0}^{x}}}f(t)~{\text{d}}t^{n-1}$

by Product Rule, Binomial formula, combination formulas, and Fundamental Theorem of Calculus I, which implies

${\underset {n~{\text{times}}}{\int \cdots \int _{0}^{x}}}f(t)~{\text{d}}t^{n}={\frac {1}{(n-1)!}}\sum _{i=1}^{n}(-1)^{i}{\binom {n-1}{i}}\left[x^{n-1-i}\int _{0}^{x}t^{i}f(t)~{\text{d}}t\right]$

and we are done. Intuitively, terms in the derivative of the form $x^{n-1}f(x)$ come from differentiating each integral (but not the power in front of it, thereby obtaining the right side $fg'$in $(fg)'=f'g+fg'$ contributing from each integral term), and cancel out by means of $0=0^{n-1}={\big (}1+(-1){\big )}^{n-1}=\sum _{i=1}^{n}(-1)^{i}{\binom {n-1}{i}}$; but terms of the form $x^{n-i-1}\int _{0}^{x}t^{i}f(t)~{\text{d}}t$ are few, with only one contribution from each integral term.

So to compute higher-order integrals, no other nonelementary integrals need to be considered except for those possibly equal to $\int _{0}^{x}t^{n}f(t)~{\text{d}}t$ for $f\in {\mathcal {R}},~n\in \mathbb {N}$ (Riemann-integrable functions). To compute from another bound $c\neq 0$, it is sufficient for integrals only of the form $\int _{0}^{x}(t-c)^{n}f(t)~{\text{d}}t$ to be considered.

Example

Below, the first five antiderivatives of $f(x)=e^{x^{2}+1/x^{2}}$ are computed and graphed, using this method. Since $f$ is undefined at $x=0$ (in fact, $x=0$ is an essential singularity, and every antiderivative diverges as $x\rightarrow 0$), we start at $x=1$ instead (a critical minimum point for $f$, so most suitable):

• $e^{x^{2}+1/x^{2}}$
• $\int _{1}^{x}e^{t^{2}+1/t^{2}}~{\text{d}}t$
• $\iint _{1}^{x}e^{t^{2}+1/t^{2}}~{\text{d}}t^{2}=(x-1)\int _{1}^{x}e^{t^{2}+1/t^{2}}~{\text{d}}t-\int _{1}^{x}(t-1)e^{t^{2}+1/t^{2}}~{\text{d}}t$
• $\iiint _{1}^{x}e^{t^{2}+1/t^{2}}~{\text{d}}t^{3}={\frac {1}{2}}\left[(x-1)^{2}\int _{1}^{x}e^{t^{2}+1/t^{2}}~{\text{d}}t-2(x-1)\int _{1}^{x}(t-1)e^{t^{2}+1/t^{2}}~{\text{d}}t+\int _{1}^{x}(t-1)^{2}e^{t^{2}+1/t^{2}}~{\text{d}}t\right]$
• ${\underset {4~{\text{times}}}{\int \cdots \int _{1}^{x}}}e^{t^{2}+1/t^{2}}~{\text{d}}t^{4}={\frac {1}{6}}\left[(x-1)^{3}\int _{1}^{x}e^{t^{2}+1/t^{2}}~{\text{d}}t-3(x-1)^{2}\int _{1}^{x}(t-1)e^{t^{2}+1/t^{2}}~{\text{d}}t+3(x-1)\int _{1}^{x}(t-1)^{2}e^{t^{2}+1/t^{2}}~{\text{d}}t-\int _{1}^{x}(t-1)^{3}e^{t^{2}+1/t^{2}}~{\text{d}}t\right]$
• ${\underset {5~{\text{times}}}{\int \cdots \int _{1}^{x}}}e^{t^{2}+1/t^{2}}~{\text{d}}t^{5}={\frac {1}{24}}\left[(x-1)^{4}\int _{1}^{x}e^{t^{2}+1/t^{2}}~{\text{d}}t-4(x-1)^{3}\int _{1}^{x}(t-1)e^{t^{2}+1/t^{2}}~{\text{d}}t+6(x-1)^{2}\int _{1}^{x}(t-1)^{2}e^{t^{2}+1/t^{2}}~{\text{d}}t-4(x-1)\int _{1}^{x}(t-1)^{3}e^{t^{2}+1/t^{2}}~{\text{d}}t+\int _{1}^{x}(t-1)^{4}e^{t^{2}+1/t^{2}}~{\text{d}}t\right]$

Note that the general chain integral formula cannot be used since (1) $g(x)=x^{2}+1/x^{2}$ is not strictly monotone, (2) no balanced essential singularity can be isolated from $e^{x^{2}+1/x^{2}}$, not even $e^{1/x^{2}}$, and (3) even the balanced essential singularity $e^{1/x^{2}}$ is not elementary-integrable.

Graphs of Five Antiderivatives of $e^{x^{2}+1/x^{2}}$

Square Window	Vertical Rectangular Window
[-9,9] × [-3.478, 10.432]	[-3.3,3.3] × [-50, 50]

Cycloid

The work-energy formula is intrinsic to any twice continuously differentiable increasing function:

${\frac {y'(x|_{y=b})^{2}-y'(x|_{y=a})^{2}}{2}}=\int _{y=a}^{y=b}y''(x|_{y})~{\text{d}}y$

This is by Chain Rule, since ${\frac {{\text{d}}f'(f^{-1}(x))^{2}}{2~{\text{d}}x}}=f''(f^{-1}(x))$ for any twice continuously differentiable increasing function $f$.

Using the work-energy formula

${\text{KE}}={\frac {1}{2}}m(v_{f}^{2}-v_{0}^{2})=\int _{x_{1}}^{x_{2}}F(x)~{\text{d}}x$

with $m=1$, initial position $x_{1}=r$, and initial velocity $v_{0}(x_{1})=0$; we have

$\left({\frac {{\text{d}}y}{{\text{d}}x}}\right)^{2}={\frac {2r}{y}}-1=v^{2}=2\int _{r}^{y}F(x)~{\text{d}}x\Longrightarrow {\frac {\text{d}}{{\text{d}}y}}\left[{\frac {2r}{y}}-1\right]={\frac {\text{d}}{{\text{d}}y}}\left[2\int _{r}^{y}F(x)~{\text{d}}x\right]\Longleftrightarrow -{\frac {2r}{y^{2}}}=2F(y)\Longrightarrow F(y)=-{\frac {r}{y^{2}}}$

by the Fundamental Theorem of Calculus I. So in fact the cycloid is the solution ${\big (}x,y(x){\big )}$ to Newton's Gravitational Law (where time is measured in $x$) if a particle is bounded in a heavy point object's gravitational field (negative net energy), with zero angular momentum; if we set $r=GM$, $y=R$, to obtain $F(R)=-{\frac {GMm}{R^{2}}}\Longrightarrow a(R)=-{\frac {GM}{R^{2}}}$.

Other zero-angular-momentum solutions for identical mass include $y(x)=x^{\frac {2}{3}}{\sqrt[{3}]{\frac {9r}{2}}}$ for zero net energy, and ${\big (}x,y(x){\big )}={\big (}r(\sinh t-t),r(\cosh t-1){\big )}$ for positive net energy. For the first case, indeed

$\left({\frac {{\text{d}}y}{{\text{d}}x}}\right)^{2}=\left[{\frac {\text{d}}{{\text{d}}x}}\left(x^{\frac {2}{3}}{\sqrt[{3}]{\frac {9r}{2}}}\right)\right]^{2}=\left[{\frac {2}{3}}{\sqrt[{3}]{\frac {9r}{2x}}}\right]^{2}={\frac {4}{9}}{\sqrt[{3}]{\frac {81r^{2}}{4x^{2}}}}={\frac {2r}{\sqrt[{3}]{{\frac {9^{3}}{81}}\cdot {\frac {4}{2^{3}}}\cdot r}}}={\frac {2r}{{\sqrt[{3}]{\frac {9r}{2}}}x^{\frac {2}{3}}}}={\frac {2r}{y}}$

by the Power Rule, where the additional constant of $-1$ is specific to the cycloid only (which is the total energy); and for the second case, indeed

$\left({\frac {{\text{d}}y}{{\text{d}}x}}\right)^{2}=\left[{\frac {{\frac {\text{d}}{{\text{d}}t}}{\big (}r(\cosh t-1){\big )}}{{\frac {\text{d}}{{\text{d}}t}}{\big (}r(\sinh t-t){\big )}}}\right]^{2}=\left[{\frac {r\sinh t}{r(\cosh t-1)}}\right]^{2}=\left[{\frac {\sinh t}{\cosh t-1}}\right]^{2}={\frac {\sinh ^{2}t}{\cosh ^{2}t-2\cosh t+1}}={\frac {\cosh ^{2}t-1}{\cosh ^{2}t-2\cosh t+1}}$

$={\frac {(\cosh t-1)(\cosh t+1)}{(\cosh t-1)^{2}}}={\frac {\cosh t+1}{\cosh t-1}}={\frac {2}{\cosh t-1}}+1={\frac {2r}{y}}+1$

by the parametric derivative, and the additional constant of $+1$ is specific to this second path only (which is the total energy).