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
edit
See Fresnel integral
We could approximate the tangent and secant integrals
T
(
x
)
:=
∫
0
x
tan
(
t
2
)
d
t
{\displaystyle T(x):=\int _{0}^{x}\tan {\big (}t^{2}{\big )}~{\text{d}}t}
Sc
(
x
)
:=
∫
0
x
sec
(
t
2
)
d
t
{\displaystyle {\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
)
=
ln
|
sec
x
2
|
2
x
+
∫
0
x
ln
|
sec
t
2
|
d
t
2
t
2
{\displaystyle T(x)={\frac {\ln |\sec x^{2}|}{2x}}+\int _{0}^{x}{\frac {\ln |\sec t^{2}|~{\text{d}}t}{2t^{2}}}}
Sc
(
x
)
=
ln
|
tan
x
2
+
sec
x
2
|
2
x
+
∫
0
x
ln
|
tan
t
2
+
sec
t
2
|
d
t
2
t
2
{\displaystyle {\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
∫
0
x
f
(
t
2
)
d
t
=
F
(
x
2
)
2
x
+
∫
0
x
F
(
t
2
)
d
t
2
x
2
{\displaystyle \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}}}}
F
(
x
)
=
∫
0
x
f
(
t
)
d
t
{\displaystyle F(x)=\int _{0}^{x}f(t)~{\text{d}}t}
In fact, integrals through CPV are defined when subtracting balanced pole functions - pure powers of
1
/
x
{\displaystyle 1/x}
O
(
n
−
α
)
{\displaystyle O(n^{-\alpha })}
α
<
1
{\displaystyle \alpha <1}
Notice that both
tan
x
2
{\displaystyle \tan x^{2}}
sec
x
2
{\displaystyle \sec x^{2}}
[
∓
1
/
x
]
x
=
0
{\displaystyle [\mp 1/x]_{x=0}}
∫
1
x
∓
d
t
t
=
∓
ln
x
{\displaystyle \int _{1}^{x}\mp {\frac {{\text{d}}t}{t}}=\mp \ln x}
∫
0
x
∓
ln
t
d
t
=
∓
x
(
ln
x
−
1
)
{\displaystyle \int _{0}^{x}\mp \ln t~{\text{d}}t=\mp x(\ln x-1)}
lim
x
→
0
∓
[
x
(
ln
x
−
1
)
]
=
0
{\displaystyle \lim _{x\rightarrow 0}\mp {\big [}x(\ln x-1){\big ]}=0}
Therefore,
∫
0
x
F
(
t
2
)
d
t
2
x
2
{\displaystyle {\frac {\int _{0}^{x}F{\big (}t^{2}{\big )}~{\text{d}}t}{2x^{2}}}}
F
(
x
)
=
ln
|
sec
x
|
=
∫
0
x
tan
t
d
t
{\displaystyle F(x)=\ln |\sec x|=\int _{0}^{x}\tan t~{\text{d}}t}
F
(
x
)
=
ln
|
tan
x
+
sec
x
|
=
∫
0
x
sec
t
d
t
{\displaystyle F(x)=\ln |\tan x+\sec x|=\int _{0}^{x}\sec t~{\text{d}}t}
∫
0
1
∓
ln
x
d
x
=
∓
[
x
(
ln
x
−
1
)
]
0
1
=
±
1
{\displaystyle \int _{0}^{1}\mp \ln x~{\text{d}}x=\mp [x(\ln x-1)]_{0}^{1}=\pm 1}
comparisons . So by CPV , the above integrals are defined except at isolated poles. Graphs of these integrals for
x
≥
0
{\displaystyle x\geq 0}
Graphs of
T
(
x
)
{\displaystyle T(x)}
Sc
(
x
)
{\displaystyle {\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:
lim inf
x
→
∞
∫
0
x
tan
(
t
2
)
d
t
=
lim
x
→
∞
∫
0
x
ln
|
sec
t
2
|
d
t
2
t
2
≈
0.535914345467
{\displaystyle \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}
lim inflection
x
→
∞
∫
0
x
sec
(
t
2
)
d
t
=
lim
x
→
∞
∫
0
x
ln
|
tan
t
2
+
sec
t
2
|
d
t
2
t
2
≈
0.836834795146
{\displaystyle {\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
π
/
8
≈
0.62665706866
{\displaystyle {\sqrt {\pi /8}}\approx 0.62665706866}
S
(
x
)
{\displaystyle S(x)}
C
(
x
)
{\displaystyle C(x)}
Modulated Integrals
edit
We could also approximate the cotangent and cosecant integrals
Ct
(
x
)
:=
∫
0
x
cot
(
t
2
)
d
t
{\displaystyle {\text{Ct}}(x):=\int _{0}^{x}\cot {\big (}t^{2}{\big )}~{\text{d}}t}
Cc
(
x
)
:=
∫
0
x
csc
(
t
2
)
d
t
{\displaystyle {\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
{\displaystyle x=0}
1
/
x
2
{\displaystyle 1/x^{2}}
f
(
x
)
{\displaystyle f(x)}
x
=
0
{\displaystyle x=0}
lim
x
→
0
f
(
x
)
=
0
{\displaystyle \lim _{x\rightarrow 0}f(x)=0}
∫
0
x
f
(
t
2
)
d
t
=
F
(
x
2
)
2
x
+
∫
0
x
F
(
t
2
)
d
t
2
x
2
{\displaystyle \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}}}}
−
1
/
x
{\displaystyle -1/x}
1
/
x
2
{\displaystyle 1/x^{2}}
Since both integrals have right limit
−
∞
{\displaystyle -\infty }
x
{\displaystyle x}
0
{\displaystyle 0}
y
=
0
{\displaystyle y=0}
x
→
∞
{\displaystyle x\rightarrow \infty }
Ct
(
x
)
:≈
1.29414
−
(
ln
2
−
ln
(
|
cot
x
2
+
csc
x
2
|
)
−
2
ln
(
|
x
|
)
2
x
2
+
1
x
+
∫
0
x
ln
2
−
ln
(
|
cot
t
2
+
csc
t
2
|
)
−
2
ln
(
|
t
|
)
2
t
2
d
t
)
{\displaystyle {\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)}
Cc
(
x
)
:≈
ln
2
−
ln
(
|
cot
x
2
+
csc
x
2
|
)
−
2
ln
(
|
x
|
)
2
x
−
1
x
+
∫
0
x
ln
2
−
ln
(
|
cot
t
2
+
csc
t
2
|
)
−
2
ln
(
|
t
|
)
2
t
2
d
t
+
0.5360919
{\displaystyle {\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
Ct
(
x
)
{\displaystyle {\text{Ct}}(x)}
Cc
(
x
)
{\displaystyle {\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
edit
A general formula for the second-order real integral (second antiderivative ) is
∬
0
x
f
(
t
)
d
t
2
=
x
∫
0
x
f
(
t
)
d
t
−
∫
0
x
t
f
(
t
)
d
t
{\displaystyle \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
d
d
x
(
x
∫
0
x
f
(
t
)
d
t
−
∫
0
x
t
f
(
t
)
d
t
)
=
x
f
(
x
)
+
∫
0
x
f
(
t
)
d
t
−
x
f
(
x
)
=
∫
0
x
f
(
t
)
d
t
;
{\displaystyle {\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;}
Product Rule and Fundamental Theorem of Calculus I .
A general formula for the third-order real integral (third antiderivative ) is
∭
0
x
f
(
t
)
d
t
3
=
1
2
(
x
2
∫
0
x
f
(
t
)
d
t
−
2
x
∫
0
x
t
f
(
t
)
d
t
+
∫
0
x
t
2
f
(
t
)
d
t
)
{\displaystyle \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
d
d
x
(
1
2
(
x
2
∫
0
x
f
(
t
)
d
t
−
2
x
∫
0
x
t
f
(
t
)
d
t
+
∫
0
x
t
2
f
(
t
)
d
t
)
)
=
1
2
(
x
2
f
(
x
)
+
2
x
∫
0
x
f
(
t
)
d
t
−
2
x
2
f
(
x
)
−
2
∫
0
x
t
f
(
t
)
d
t
+
x
2
f
(
x
)
)
=
1
2
(
2
x
∫
0
x
f
(
t
)
d
t
−
2
∫
0
x
t
f
(
t
)
d
t
)
=
∬
0
x
f
(
t
)
d
t
2
;
{\displaystyle {\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};}
Product Rule and Fundamental Theorem of Calculus I .
A general formula for the
n
{\displaystyle n}
real integral (
n
{\displaystyle n}
antiderivative ) is
∫
⋯
∫
0
x
n
times
f
(
t
)
d
t
n
=
1
(
n
−
1
)
!
∑
i
=
1
n
(
−
1
)
i
(
n
−
1
i
)
[
x
n
−
1
−
i
∫
0
x
t
i
f
(
t
)
d
t
]
{\displaystyle {\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
∫
⋯
∫
0
x
n
−
1
times
f
(
t
)
d
t
n
−
1
=
1
(
n
−
2
)
!
∑
i
=
1
n
−
1
(
−
1
)
i
(
n
−
2
i
)
[
x
n
−
2
−
i
∫
0
x
t
i
f
(
t
)
d
t
]
{\displaystyle {\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,
d
d
x
(
1
(
n
−
1
)
!
∑
i
=
1
n
(
−
1
)
i
(
n
−
1
i
)
[
x
n
−
i
−
1
∫
0
x
t
i
f
(
t
)
d
t
]
)
=
1
(
n
−
1
)
!
∑
i
=
1
n
(
−
1
)
i
(
n
−
1
i
)
[
x
n
−
i
−
1
x
i
f
(
x
)
+
(
n
−
i
−
1
)
x
n
−
i
−
2
∫
0
x
t
i
f
(
t
)
d
t
]
{\displaystyle {\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]}
=
1
(
n
−
1
)
!
[
∑
i
=
1
n
(
−
1
)
i
(
n
−
1
i
)
x
n
−
1
f
(
x
)
+
∑
i
=
1
n
(
−
1
)
n
(
n
−
1
i
)
∫
0
x
t
i
f
(
t
)
d
t
]
=
1
(
n
−
1
)
!
[
x
n
−
1
f
(
x
)
(
1
+
(
−
1
)
)
n
−
1
+
∑
i
=
1
n
(
−
1
)
n
(
n
−
1
)
!
i
!
(
n
−
i
−
1
)
!
(
n
−
i
−
1
)
x
n
−
i
−
2
∫
0
x
t
i
f
(
t
)
d
t
]
{\displaystyle ={\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]}
=
1
(
n
−
1
)
!
[
x
n
−
1
f
(
x
)
⋅
0
n
−
1
+
∑
i
=
1
n
(
−
1
)
n
(
n
−
1
)
!
i
!
(
n
−
i
−
2
)
!
x
n
−
i
−
2
∫
0
x
t
i
f
(
t
)
d
t
]
=
1
(
n
−
1
)
!
[
(
n
−
1
)
∑
i
=
1
n
(
−
1
)
n
(
n
−
2
)
!
i
!
(
n
−
i
−
2
)
!
x
n
−
i
−
2
∫
0
x
t
i
f
(
t
)
d
t
]
{\displaystyle ={\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]}
=
1
(
n
−
2
)
!
∑
i
=
1
n
−
1
(
−
1
)
i
(
n
−
2
i
)
[
x
n
−
2
−
i
∫
0
x
t
i
f
(
t
)
d
t
]
=
∫
⋯
∫
0
x
n
−
1
times
f
(
t
)
d
t
n
−
1
{\displaystyle ={\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
∫
⋯
∫
0
x
n
times
f
(
t
)
d
t
n
=
1
(
n
−
1
)
!
∑
i
=
1
n
(
−
1
)
i
(
n
−
1
i
)
[
x
n
−
1
−
i
∫
0
x
t
i
f
(
t
)
d
t
]
{\displaystyle {\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
)
{\displaystyle x^{n-1}f(x)}
f
g
′
{\displaystyle fg'}
(
f
g
)
′
=
f
′
g
+
f
g
′
{\displaystyle (fg)'=f'g+fg'}
0
=
0
n
−
1
=
(
1
+
(
−
1
)
)
n
−
1
=
∑
i
=
1
n
(
−
1
)
i
(
n
−
1
i
)
{\displaystyle 0=0^{n-1}={\big (}1+(-1){\big )}^{n-1}=\sum _{i=1}^{n}(-1)^{i}{\binom {n-1}{i}}}
x
n
−
i
−
1
∫
0
x
t
i
f
(
t
)
d
t
{\displaystyle x^{n-i-1}\int _{0}^{x}t^{i}f(t)~{\text{d}}t}
So to compute higher-order integrals, no other nonelementary integrals need to be considered except for those possibly equal to
∫
0
x
t
n
f
(
t
)
d
t
{\displaystyle \int _{0}^{x}t^{n}f(t)~{\text{d}}t}
f
∈
R
,
n
∈
N
{\displaystyle f\in {\mathcal {R}},~n\in \mathbb {N} }
Riemann-integrable functions). To compute from another bound
c
≠
0
{\displaystyle c\neq 0}
only of the form
∫
0
x
(
t
−
c
)
n
f
(
t
)
d
t
{\displaystyle \int _{0}^{x}(t-c)^{n}f(t)~{\text{d}}t}
Below, the first five antiderivatives of
f
(
x
)
=
e
x
2
+
1
/
x
2
{\displaystyle f(x)=e^{x^{2}+1/x^{2}}}
f
{\displaystyle f}
x
=
0
{\displaystyle x=0}
x
=
0
{\displaystyle x=0}
essential singularity , and every antiderivative diverges as
x
→
0
{\displaystyle x\rightarrow 0}
x
=
1
{\displaystyle x=1}
critical minimum point for
f
{\displaystyle f}
e
x
2
+
1
/
x
2
{\displaystyle e^{x^{2}+1/x^{2}}}
∫
1
x
e
t
2
+
1
/
t
2
d
t
{\displaystyle \int _{1}^{x}e^{t^{2}+1/t^{2}}~{\text{d}}t}
∬
1
x
e
t
2
+
1
/
t
2
d
t
2
=
(
x
−
1
)
∫
1
x
e
t
2
+
1
/
t
2
d
t
−
∫
1
x
(
t
−
1
)
e
t
2
+
1
/
t
2
d
t
{\displaystyle \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}
∭
1
x
e
t
2
+
1
/
t
2
d
t
3
=
1
2
[
(
x
−
1
)
2
∫
1
x
e
t
2
+
1
/
t
2
d
t
−
2
(
x
−
1
)
∫
1
x
(
t
−
1
)
e
t
2
+
1
/
t
2
d
t
+
∫
1
x
(
t
−
1
)
2
e
t
2
+
1
/
t
2
d
t
]
{\displaystyle \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]}
∫
⋯
∫
1
x
4
times
e
t
2
+
1
/
t
2
d
t
4
=
1
6
[
(
x
−
1
)
3
∫
1
x
e
t
2
+
1
/
t
2
d
t
−
3
(
x
−
1
)
2
∫
1
x
(
t
−
1
)
e
t
2
+
1
/
t
2
d
t
+
3
(
x
−
1
)
∫
1
x
(
t
−
1
)
2
e
t
2
+
1
/
t
2
d
t
−
∫
1
x
(
t
−
1
)
3
e
t
2
+
1
/
t
2
d
t
]
{\displaystyle {\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]}
∫
⋯
∫
1
x
5
times
e
t
2
+
1
/
t
2
d
t
5
=
1
24
[
(
x
−
1
)
4
∫
1
x
e
t
2
+
1
/
t
2
d
t
−
4
(
x
−
1
)
3
∫
1
x
(
t
−
1
)
e
t
2
+
1
/
t
2
d
t
+
6
(
x
−
1
)
2
∫
1
x
(
t
−
1
)
2
e
t
2
+
1
/
t
2
d
t
−
4
(
x
−
1
)
∫
1
x
(
t
−
1
)
3
e
t
2
+
1
/
t
2
d
t
+
∫
1
x
(
t
−
1
)
4
e
t
2
+
1
/
t
2
d
t
]
{\displaystyle {\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
{\displaystyle g(x)=x^{2}+1/x^{2}}
e
x
2
+
1
/
x
2
{\displaystyle e^{x^{2}+1/x^{2}}}
e
1
/
x
2
{\displaystyle e^{1/x^{2}}}
e
1
/
x
2
{\displaystyle e^{1/x^{2}}}
The work-energy formula is intrinsic to any twice continuously differentiable increasing function:
y
′
(
x
|
y
=
b
)
2
−
y
′
(
x
|
y
=
a
)
2
2
=
∫
y
=
a
y
=
b
y
″
(
x
|
y
)
d
y
{\displaystyle {\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
d
f
′
(
f
−
1
(
x
)
)
2
2
d
x
=
f
″
(
f
−
1
(
x
)
)
{\displaystyle {\frac {{\text{d}}f'(f^{-1}(x))^{2}}{2~{\text{d}}x}}=f''(f^{-1}(x))}
f
{\displaystyle f}
Using the work-energy formula
KE
=
1
2
m
(
v
f
2
−
v
0
2
)
=
∫
x
1
x
2
F
(
x
)
d
x
{\displaystyle {\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
{\displaystyle m=1}
x
1
=
r
{\displaystyle x_{1}=r}
v
0
(
x
1
)
=
0
{\displaystyle v_{0}(x_{1})=0}
(
d
y
d
x
)
2
=
2
r
y
−
1
=
v
2
=
2
∫
r
y
F
(
x
)
d
x
⟹
d
d
y
[
2
r
y
−
1
]
=
d
d
y
[
2
∫
r
y
F
(
x
)
d
x
]
⟺
−
2
r
y
2
=
2
F
(
y
)
⟹
F
(
y
)
=
−
r
y
2
{\displaystyle \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
(
x
,
y
(
x
)
)
{\displaystyle {\big (}x,y(x){\big )}}
Newton's Gravitational Law (where time is measured in
x
{\displaystyle x}
point object's gravitational field (negative net energy), with zero angular momentum; if we set
r
=
G
M
{\displaystyle r=GM}
y
=
R
{\displaystyle y=R}
F
(
R
)
=
−
G
M
m
R
2
⟹
a
(
R
)
=
−
G
M
R
2
{\displaystyle 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
2
3
9
r
2
3
{\displaystyle y(x)=x^{\frac {2}{3}}{\sqrt[{3}]{\frac {9r}{2}}}}
(
x
,
y
(
x
)
)
=
(
r
(
sinh
t
−
t
)
,
r
(
cosh
t
−
1
)
)
{\displaystyle {\big (}x,y(x){\big )}={\big (}r(\sinh t-t),r(\cosh t-1){\big )}}
(
d
y
d
x
)
2
=
[
d
d
x
(
x
2
3
9
r
2
3
)
]
2
=
[
2
3
9
r
2
x
3
]
2
=
4
9
81
r
2
4
x
2
3
=
2
r
9
3
81
⋅
4
2
3
⋅
r
3
=
2
r
9
r
2
3
x
2
3
=
2
r
y
{\displaystyle \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
{\displaystyle -1}
(
d
y
d
x
)
2
=
[
d
d
t
(
r
(
cosh
t
−
1
)
)
d
d
t
(
r
(
sinh
t
−
t
)
)
]
2
=
[
r
sinh
t
r
(
cosh
t
−
1
)
]
2
=
[
sinh
t
cosh
t
−
1
]
2
=
sinh
2
t
cosh
2
t
−
2
cosh
t
+
1
=
cosh
2
t
−
1
cosh
2
t
−
2
cosh
t
+
1
{\displaystyle \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}}}
=
(
cosh
t
−
1
)
(
cosh
t
+
1
)
(
cosh
t
−
1
)
2
=
cosh
t
+
1
cosh
t
−
1
=
2
cosh
t
−
1
+
1
=
2
r
y
+
1
{\displaystyle ={\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
{\displaystyle +1}
![{\displaystyle {\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/919f14dbe098ed371b1cd9c3835c01b6a2504732)
![{\displaystyle [\mp 1/x]_{x=0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/906870be0023f09a68d48c43ff248bd8967e78aa)
![{\displaystyle \lim _{x\rightarrow 0}\mp {\big [}x(\ln x-1){\big ]}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c89c8d7ff5170dcf1163eb2bd7fc9495bf71555a)
![{\displaystyle \int _{0}^{1}\mp \ln x~{\text{d}}x=\mp [x(\ln x-1)]_{0}^{1}=\pm 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3bb058105caaf962e28c992ef5acf1e6c77cb76c)
![{\displaystyle {\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7e29cebd9977af9f04fa59418d555bb06409fed)
![{\displaystyle {\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/887c55e2f43682187784a50267075fc08417661f)
![{\displaystyle {\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e8afd39287c0a059b0e36aefa3c6c21e1f3e5c5)
![{\displaystyle ={\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/97038adcdedee6156285e61bf3b370107cc3aa67)
![{\displaystyle ={\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d1f96128e32135f5cd2733c8d61d2cf39c6bf07)
![{\displaystyle ={\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/672c00914d73000b10a84638e17f0d37589ff9a7)
![{\displaystyle \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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1e0bce45e1402d7ef06e7d04a8e8e6887a2ccc6)
![{\displaystyle {\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f825a93e5ad5d4b01e53711283ff5efe25043e72)
![{\displaystyle {\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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf79a6759b175e85f89edbe531e9e2fbcefdd66e)
).png)
)%20II.png)
![{\displaystyle \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}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/185e4b0e67e742bcd0847f8ce357a2dbaaaa29b7)
![{\displaystyle y(x)=x^{\frac {2}{3}}{\sqrt[{3}]{\frac {9r}{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f15f53bd3347a2e3efa761118658f40b817a17b)
![{\displaystyle \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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa33392c81a3a5f905e5f691ed996946ed1dd41d)
![{\displaystyle \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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f2e65cb794c3b23aeef6950549d13db938e6648)
