S (x ) and C (x ) The maximum of C (x ) is about 0.977451424. If πt ²/2 were used instead of t ², then the image would be scaled vertically and horizontally (see below). Credit: .
Normalised Fresnel integrals, S (x ) and C (x ) have the argument of the trigonometric function is πt 2 /2, as opposed to just t 2 as above. Credit: .
For any real value of the argument
x
{\displaystyle x}
, the Fresnel integrals
C
(
x
)
{\displaystyle C(x)}
and
S
(
x
)
{\displaystyle S(x)}
are defined as the integrals:
C
(
x
)
=
∫
0
x
cos
t
2
d
t
,
{\displaystyle C(x)\;=\;\int _{0}^{x}\cos {t^{2}}\,dt,}
and
S
(
x
)
=
∫
0
x
sin
t
2
d
t
.
{\displaystyle S(x)\;=\;\int _{0}^{x}\sin {t^{2}}\,dt.}
In optics, both of them express the intensity of diffracted light behind an illuminated edge.
Using the Taylor series expansions of cosine and sine, we get easily the expansions of the functions :
C
(
z
)
=
z
1
−
z
5
5
⋅
2
!
+
z
9
9
⋅
4
!
−
z
13
13
⋅
6
!
+
−
…
{\displaystyle C(z)\,=\,{\frac {z}{1}}\!-\!{\frac {z^{5}}{5\!\cdot \!2!}}\!+\!{\frac {z^{9}}{9\!\cdot \!4!}}\!-\!{\frac {z^{13}}{13\!\cdot \!6!}}\!+\!-\ldots }
S
(
z
)
=
z
3
3
⋅
1
!
−
z
7
7
⋅
3
!
+
z
11
11
⋅
5
!
−
z
15
15
⋅
7
!
+
−
…
{\displaystyle S(z)\,=\,{\frac {z^{3}}{3\cdot 1!}}\!-\!{\frac {z^{7}}{7\!\cdot \!3!}}\!+\!{\frac {z^{11}}{11\!\cdot \!5!}}\!-\!{\frac {z^{15}}{15\!\cdot \!7!}}\!+\!-\ldots }
S
(
z
)
=
∫
0
z
sin
(
t
2
)
d
t
=
∑
n
=
0
∞
(
−
1
)
n
z
4
n
+
3
(
2
n
+
1
)
!
(
4
n
+
3
)
{\displaystyle S(z)=\int _{0}^{z}\sin(t^{2})\,\mathrm {d} t=\sum _{n=0}^{\infty }(-1)^{n}{\frac {z^{4n+3}}{(2n+1)!(4n+3)}}}
C
(
z
)
=
∫
0
z
cos
(
t
2
)
d
t
=
∑
n
=
0
∞
(
−
1
)
n
z
4
n
+
1
(
2
n
)
!
(
4
n
+
1
)
{\displaystyle C(z)=\int _{0}^{z}\cos(t^{2})\,\mathrm {d} t=\sum _{n=0}^{\infty }(-1)^{n}{\frac {z^{4n+1}}{(2n)!(4n+1)}}}
These converge for all complex values
z
{\displaystyle z}
, and thus define entire transcendental functions.
The Fresnel integrals at infinity have the finite value
lim
x
→
∞
C
(
x
)
=
lim
x
→
∞
S
(
x
)
=
2
π
4
.
{\displaystyle \lim _{x\to \infty }C(x)=\lim _{x\to \infty }S(x)={\frac {\sqrt {2\pi }}{4}}.}
The parametric presentation
x
=
C
(
t
)
,
y
=
S
(
t
)
{\displaystyle {\begin{matrix}x\,=\,C(t),\quad y=S(t)\end{matrix}}}
represents a curve called clothoid .
Since the equations both define odd functions, the clothoid has symmetry about the origin.
The curve has the shape of a "
∼
{\displaystyle \sim }
"
(see this diagram ).
The arc length of the clothoid from the origin to the point ,
(
C
(
t
)
,
S
(
t
)
)
{\displaystyle (C(t),\,S(t))}
, is simply
∫
0
t
C
′
(
u
)
2
+
S
′
(
u
)
2
d
u
=
∫
0
t
cos
2
(
u
2
)
+
sin
2
(
u
2
)
d
u
=
∫
0
t
d
u
=
t
.
{\displaystyle \int _{0}^{t}{\sqrt {C'(u)^{2}+S'(u)^{2}}}\,du=\int _{0}^{t}{\sqrt {\cos ^{2}(u^{2})+\sin ^{2}(u^{2})}}\,du=\int _{0}^{t}du=t.}
Thus, the length of the whole curve to the point,
(
2
π
4
,
2
π
4
)
{\displaystyle ({\frac {\sqrt {2\pi }}{4}},\,{\frac {\sqrt {2\pi }}{4}})}
is infinite.
The curvature of the clothoid also is extremely simple,
ϰ
=
2
t
,
{\displaystyle \varkappa \,=\,2t,}
i.e. proportional to the arc lenth; thus in the origin only the curvature is zero.
Conversely, if the curvature of a plane curve varies proportionally to the arc length, the curve is a clothoid.
This property of the curvature of clothoid is utilised in way and railway construction, since the form of the clothoid is very efficient when a straight portion of way must be bent to a turn, the zero curvature of the line can be continuously raised to the wished curvature.