Let
f
(
t
)
{\displaystyle f(t)}
be a function defined whenever
t
≥
0.
{\displaystyle t\geq 0.}
Then its Laplace transform is given by
L
{
f
(
t
)
}
=
F
(
s
)
=
∫
0
∞
e
−
s
t
f
(
t
)
d
t
,
{\displaystyle {\mathcal {L}}\{f(t)\}=F(s)=\int _{0}^{\infty }e^{-st}f(t)\,dt,}
if the integral exists.[ 3]
A property of the Laplace transform useful for evaluating improper integrals is
L
[
f
(
t
)
t
]
=
∫
s
∞
F
(
u
)
d
u
,
{\displaystyle {\mathcal {L}}\left[{\frac {f(t)}{t}}\right]=\int _{s}^{\infty }F(u)\,du,}
provided
lim
t
→
0
f
(
t
)
t
{\displaystyle \lim _{t\to 0}{\frac {f(t)}{t}}}
exists.
In what follows, one needs the result
L
{
sin
t
}
=
1
s
2
+
1
,
{\displaystyle {\mathcal {L}}\{\sin t\}={\frac {1}{s^{2}+1}},}
which is the Laplace transform of the function
sin
t
{\displaystyle \sin t}
(see the section 'Differentiating under the integral sign' for a derivation) as well as a version of Abel's theorem (a consequence of the final value theorem for the Laplace transform ).
Therefore,
∫
0
∞
sin
t
t
d
t
=
lim
s
→
0
∫
0
∞
e
−
s
t
sin
t
t
d
t
=
lim
s
→
0
L
[
sin
t
t
]
=
lim
s
→
0
∫
s
∞
d
u
u
2
+
1
=
lim
s
→
0
arctan
u
|
s
∞
=
lim
s
→
0
[
π
2
−
arctan
(
s
)
]
=
π
2
.
{\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {\sin t}{t}}\,dt&=\lim _{s\to 0}\int _{0}^{\infty }e^{-st}{\frac {\sin t}{t}}\,dt=\lim _{s\to 0}{\mathcal {L}}\left[{\frac {\sin t}{t}}\right]\\[6pt]&=\lim _{s\to 0}\int _{s}^{\infty }{\frac {du}{u^{2}+1}}=\lim _{s\to 0}\arctan u{\Biggr |}_{s}^{\infty }\\[6pt]&=\lim _{s\to 0}\left[{\frac {\pi }{2}}-\arctan(s)\right]={\frac {\pi }{2}}.\end{aligned}}}
Evaluating the Dirichlet integral using the Laplace transform is equivalent to calculating the same double definite integral by changing the order of integration , namely,
(
I
1
=
∫
0
∞
∫
0
∞
e
−
s
t
sin
t
d
t
d
s
)
=
(
I
2
=
∫
0
∞
∫
0
∞
e
−
s
t
sin
t
d
s
d
t
)
,
{\displaystyle \left(I_{1}=\int _{0}^{\infty }\int _{0}^{\infty }e^{-st}\sin t\,dt\,ds\right)=\left(I_{2}=\int _{0}^{\infty }\int _{0}^{\infty }e^{-st}\sin t\,ds\,dt\right),}
(
I
1
=
∫
0
∞
1
s
2
+
1
d
s
=
π
2
)
=
(
I
2
=
∫
0
∞
sin
t
t
d
t
)
,
provided
s
>
0.
{\displaystyle \left(I_{1}=\int _{0}^{\infty }{\frac {1}{s^{2}+1}}\,ds={\frac {\pi }{2}}\right)=\left(I_{2}=\int _{0}^{\infty }{\frac {\sin t}{t}}\,dt\right),{\text{ provided }}s>0.}
The change of order is justified by the fact that for all
s
>
0
{\displaystyle s>0}
, the integral is absolutely convergent.
Differentiation under the integral sign (Feynman's trick)
edit
First rewrite the integral as a function of the additional variable
s
,
{\displaystyle s,}
namely, the Laplace transform of
sin
t
t
.
{\displaystyle {\frac {\sin t}{t}}.}
So let
f
(
s
)
=
∫
0
∞
e
−
s
t
sin
t
t
d
t
.
{\displaystyle f(s)=\int _{0}^{\infty }e^{-st}{\frac {\sin t}{t}}\,dt.}
In order to evaluate the Dirichlet integral, we need to determine
f
(
0
)
.
{\displaystyle f(0).}
The continuity of
f
{\displaystyle f}
can be justified by applying the dominated convergence theorem after integration by parts. Differentiate with respect to
s
>
0
{\displaystyle s>0}
and apply the Leibniz rule for differentiating under the integral sign to obtain
d
f
d
s
=
d
d
s
∫
0
∞
e
−
s
t
sin
t
t
d
t
=
∫
0
∞
∂
∂
s
e
−
s
t
sin
t
t
d
t
=
−
∫
0
∞
e
−
s
t
sin
t
d
t
.
{\displaystyle {\begin{aligned}{\frac {df}{ds}}&={\frac {d}{ds}}\int _{0}^{\infty }e^{-st}{\frac {\sin t}{t}}\,dt=\int _{0}^{\infty }{\frac {\partial }{\partial s}}e^{-st}{\frac {\sin t}{t}}\,dt\\[6pt]&=-\int _{0}^{\infty }e^{-st}\sin t\,dt.\end{aligned}}}
Now, using Euler's formula
e
i
t
=
cos
t
+
i
sin
t
,
{\displaystyle e^{it}=\cos t+i\sin t,}
one can express the sine function in terms of complex exponentials:
sin
t
=
1
2
i
(
e
i
t
−
e
−
i
t
)
.
{\displaystyle \sin t={\frac {1}{2i}}\left(e^{it}-e^{-it}\right).}
Therefore,
d
f
d
s
=
−
∫
0
∞
e
−
s
t
sin
t
d
t
=
−
∫
0
∞
e
−
s
t
e
i
t
−
e
−
i
t
2
i
d
t
=
−
1
2
i
∫
0
∞
[
e
−
t
(
s
−
i
)
−
e
−
t
(
s
+
i
)
]
d
t
=
−
1
2
i
[
−
1
s
−
i
e
−
t
(
s
−
i
)
−
−
1
s
+
i
e
−
t
(
s
+
i
)
]
0
∞
=
−
1
2
i
[
0
−
(
−
1
s
−
i
+
1
s
+
i
)
]
=
−
1
2
i
(
1
s
−
i
−
1
s
+
i
)
=
−
1
2
i
(
s
+
i
−
(
s
−
i
)
s
2
+
1
)
=
−
1
s
2
+
1
.
{\displaystyle {\begin{aligned}{\frac {df}{ds}}&=-\int _{0}^{\infty }e^{-st}\sin t\,dt=-\int _{0}^{\infty }e^{-st}{\frac {e^{it}-e^{-it}}{2i}}dt\\[6pt]&=-{\frac {1}{2i}}\int _{0}^{\infty }\left[e^{-t(s-i)}-e^{-t(s+i)}\right]dt\\[6pt]&=-{\frac {1}{2i}}\left[{\frac {-1}{s-i}}e^{-t(s-i)}-{\frac {-1}{s+i}}e^{-t(s+i)}\right]_{0}^{\infty }\\[6pt]&=-{\frac {1}{2i}}\left[0-\left({\frac {-1}{s-i}}+{\frac {1}{s+i}}\right)\right]=-{\frac {1}{2i}}\left({\frac {1}{s-i}}-{\frac {1}{s+i}}\right)\\[6pt]&=-{\frac {1}{2i}}\left({\frac {s+i-(s-i)}{s^{2}+1}}\right)=-{\frac {1}{s^{2}+1}}.\end{aligned}}}
Integrating with respect to
s
{\displaystyle s}
gives
f
(
s
)
=
∫
−
d
s
s
2
+
1
=
A
−
arctan
s
,
{\displaystyle f(s)=\int {\frac {-ds}{s^{2}+1}}=A-\arctan s,}
where
A
{\displaystyle A}
is a constant of integration to be determined. Since
lim
s
→
∞
f
(
s
)
=
0
,
{\displaystyle \lim _{s\to \infty }f(s)=0,}
A
=
lim
s
→
∞
arctan
s
=
π
2
,
{\displaystyle A=\lim _{s\to \infty }\arctan s={\frac {\pi }{2}},}
using the principal value. This means that for
s
>
0
{\displaystyle s>0}
f
(
s
)
=
π
2
−
arctan
s
.
{\displaystyle f(s)={\frac {\pi }{2}}-\arctan s.}
Finally, by continuity at
s
=
0
,
{\displaystyle s=0,}
we have
f
(
0
)
=
π
2
−
arctan
(
0
)
=
π
2
,
{\displaystyle f(0)={\frac {\pi }{2}}-\arctan(0)={\frac {\pi }{2}},}
as before.
Complex contour integration
edit
Consider
f
(
z
)
=
e
i
z
z
.
{\displaystyle f(z)={\frac {e^{iz}}{z}}.}
As a function of the complex variable
z
,
{\displaystyle z,}
it has a simple pole at the origin, which prevents the application of Jordan's lemma , whose other hypotheses are satisfied.
Define then a new function[ 4]
g
(
z
)
=
e
i
z
z
+
i
ε
.
{\displaystyle g(z)={\frac {e^{iz}}{z+i\varepsilon }}.}
The pole has been moved to the negative imaginary axis, so
g
(
z
)
{\displaystyle g(z)}
can be integrated along the semicircle
γ
{\displaystyle \gamma }
of radius
R
{\displaystyle R}
centered at
z
=
0
{\displaystyle z=0}
extending in the positive imaginary direction, and closed along the real axis. One then takes the limit
ε
→
0.
{\displaystyle \varepsilon \to 0.}
The complex integral is zero by the residue theorem , as there are no poles inside the integration path
γ
{\displaystyle \gamma }
:
0
=
∫
γ
g
(
z
)
d
z
=
∫
−
R
R
e
i
x
x
+
i
ε
d
x
+
∫
0
π
e
i
(
R
e
i
θ
+
θ
)
R
e
i
θ
+
i
ε
i
R
d
θ
.
{\displaystyle 0=\int _{\gamma }g(z)\,dz=\int _{-R}^{R}{\frac {e^{ix}}{x+i\varepsilon }}\,dx+\int _{0}^{\pi }{\frac {e^{i(Re^{i\theta }+\theta )}}{Re^{i\theta }+i\varepsilon }}iR\,d\theta .}
The second term vanishes as
R
{\displaystyle R}
goes to infinity. As for the first integral, one can use one version of the Sokhotski–Plemelj theorem for integrals over the real line: for a complex -valued function f defined and continuously differentiable on the real line and real constants
a
{\displaystyle a}
and
b
{\displaystyle b}
with
a
<
0
<
b
{\displaystyle a<0<b}
one finds
lim
ε
→
0
+
∫
a
b
f
(
x
)
x
±
i
ε
d
x
=
∓
i
π
f
(
0
)
+
P
∫
a
b
f
(
x
)
x
d
x
,
{\displaystyle \lim _{\varepsilon \to 0^{+}}\int _{a}^{b}{\frac {f(x)}{x\pm i\varepsilon }}\,dx=\mp i\pi f(0)+{\mathcal {P}}\int _{a}^{b}{\frac {f(x)}{x}}\,dx,}
where
P
{\displaystyle {\mathcal {P}}}
denotes the Cauchy principal value . Back to the above original calculation, one can write
0
=
P
∫
e
i
x
x
d
x
−
π
i
.
{\displaystyle 0={\mathcal {P}}\int {\frac {e^{ix}}{x}}\,dx-\pi i.}
By taking the imaginary part on both sides and noting that the function
sin
(
x
)
/
x
{\displaystyle \sin(x)/x}
is even, we get
∫
−
∞
+
∞
sin
(
x
)
x
d
x
=
2
∫
0
+
∞
sin
(
x
)
x
d
x
.
{\displaystyle \int _{-\infty }^{+\infty }{\frac {\sin(x)}{x}}\,dx=2\int _{0}^{+\infty }{\frac {\sin(x)}{x}}\,dx.}
Finally,
lim
ε
→
0
∫
ε
∞
sin
(
x
)
x
d
x
=
∫
0
∞
sin
(
x
)
x
d
x
=
π
2
.
{\displaystyle \lim _{\varepsilon \to 0}\int _{\varepsilon }^{\infty }{\frac {\sin(x)}{x}}\,dx=\int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}.}
Alternatively, choose as the integration contour for
f
{\displaystyle f}
the union of upper half-plane semicircles of radii
ε
{\displaystyle \varepsilon }
and
R
{\displaystyle R}
together with two segments of the real line that connect them. On one hand the contour integral is zero, independently of
ε
{\displaystyle \varepsilon }
and
R
;
{\displaystyle R;}
on the other hand, as
ε
→
0
{\displaystyle \varepsilon \to 0}
and
R
→
∞
{\displaystyle R\to \infty }
the integral's imaginary part converges to
2
I
+
ℑ
(
ln
0
−
ln
(
π
i
)
)
=
2
I
−
π
{\displaystyle 2I+\Im {\big (}\ln 0-\ln(\pi i){\big )}=2I-\pi }
(here
ln
z
{\displaystyle \ln z}
is any branch of logarithm on upper half-plane), leading to
I
=
π
2
.
{\displaystyle I={\frac {\pi }{2}}.}
Consider the well-known formula for the Dirichlet kernel :[ 5]
D
n
(
x
)
=
1
+
2
∑
k
=
1
n
cos
(
2
k
x
)
=
sin
[
(
2
n
+
1
)
x
]
sin
(
x
)
.
{\displaystyle D_{n}(x)=1+2\sum _{k=1}^{n}\cos(2kx)={\frac {\sin[(2n+1)x]}{\sin(x)}}.}
It immediately follows that:
∫
0
π
2
D
n
(
x
)
d
x
=
π
2
.
{\displaystyle \int _{0}^{\frac {\pi }{2}}D_{n}(x)\,dx={\frac {\pi }{2}}.}
Define
f
(
x
)
=
{
1
x
−
1
sin
(
x
)
x
≠
0
0
x
=
0
{\displaystyle f(x)={\begin{cases}{\frac {1}{x}}-{\frac {1}{\sin(x)}}&x\neq 0\\[6pt]0&x=0\end{cases}}}
Clearly,
f
{\displaystyle f}
is continuous when
x
∈
(
0
,
π
/
2
]
;
{\displaystyle x\in (0,\pi /2];}
to see its continuity at 0 apply L'Hopital's Rule :
lim
x
→
0
sin
(
x
)
−
x
x
sin
(
x
)
=
lim
x
→
0
cos
(
x
)
−
1
sin
(
x
)
+
x
cos
(
x
)
=
lim
x
→
0
−
sin
(
x
)
2
cos
(
x
)
−
x
sin
(
x
)
=
0.
{\displaystyle \lim _{x\to 0}{\frac {\sin(x)-x}{x\sin(x)}}=\lim _{x\to 0}{\frac {\cos(x)-1}{\sin(x)+x\cos(x)}}=\lim _{x\to 0}{\frac {-\sin(x)}{2\cos(x)-x\sin(x)}}=0.}
Hence,
f
{\displaystyle f}
fulfills the requirements of the Riemann-Lebesgue Lemma . This means:
lim
λ
→
∞
∫
0
π
/
2
f
(
x
)
sin
(
λ
x
)
d
x
=
0
⟹
lim
λ
→
∞
∫
0
π
/
2
sin
(
λ
x
)
x
d
x
=
lim
λ
→
∞
∫
0
π
/
2
sin
(
λ
x
)
sin
(
x
)
d
x
.
{\displaystyle \lim _{\lambda \to \infty }\int _{0}^{\pi /2}f(x)\sin(\lambda x)dx=0\quad \Longrightarrow \quad \lim _{\lambda \to \infty }\int _{0}^{\pi /2}{\frac {\sin(\lambda x)}{x}}dx=\lim _{\lambda \to \infty }\int _{0}^{\pi /2}{\frac {\sin(\lambda x)}{\sin(x)}}dx.}
(The form of the Riemann-Lebesgue Lemma used here is proven in the article cited.)
We would like to compute:
∫
0
∞
sin
(
t
)
t
d
t
=
lim
λ
→
∞
∫
0
λ
π
2
sin
(
t
)
t
d
t
=
lim
λ
→
∞
∫
0
π
2
sin
(
λ
x
)
x
d
x
=
lim
λ
→
∞
∫
0
π
2
sin
(
λ
x
)
sin
(
x
)
d
x
=
lim
n
→
∞
∫
0
π
2
sin
(
(
2
n
+
1
)
x
)
sin
(
x
)
d
x
=
lim
n
→
∞
∫
0
π
2
D
n
(
x
)
d
x
=
π
2
{\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {\sin(t)}{t}}dt=&\lim _{\lambda \to \infty }\int _{0}^{\lambda {\frac {\pi }{2}}}{\frac {\sin(t)}{t}}dt\\[6pt]=&\lim _{\lambda \to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin(\lambda x)}{x}}dx\\[6pt]=&\lim _{\lambda \to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin(\lambda x)}{\sin(x)}}dx\\[6pt]=&\lim _{n\to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin((2n+1)x)}{\sin(x)}}dx\\[6pt]=&\lim _{n\to \infty }\int _{0}^{\frac {\pi }{2}}D_{n}(x)dx={\frac {\pi }{2}}\end{aligned}}}
However, we must justify switching the real limit in
λ
{\displaystyle \lambda }
to the integral limit in
n
,
{\displaystyle n,}
which will follow from showing that the limit does exist.
Using integration by parts , we have:
∫
a
b
sin
(
x
)
x
d
x
=
∫
a
b
d
(
1
−
cos
(
x
)
)
x
d
x
=
1
−
cos
(
x
)
x
|
a
b
+
∫
a
b
1
−
cos
(
x
)
x
2
d
x
{\displaystyle \int _{a}^{b}{\frac {\sin(x)}{x}}dx=\int _{a}^{b}{\frac {d(1-\cos(x))}{x}}dx=\left.{\frac {1-\cos(x)}{x}}\right|_{a}^{b}+\int _{a}^{b}{\frac {1-\cos(x)}{x^{2}}}dx}
Now, as
a
→
0
{\displaystyle a\to 0}
and
b
→
∞
{\displaystyle b\to \infty }
the term on the left converges with no problem. See the list of limits of trigonometric functions . We now show that
∫
−
∞
∞
1
−
cos
(
x
)
x
2
d
x
{\displaystyle \int _{-\infty }^{\infty }{\frac {1-\cos(x)}{x^{2}}}dx}
is absolutely integrable, which implies that the limit exists.[ 6]
First, we seek to bound the integral near the origin. Using the Taylor-series expansion of the cosine about zero,
1
−
cos
(
x
)
=
1
−
∑
k
≥
0
(
−
1
)
(
k
+
1
)
x
2
k
2
k
!
=
∑
k
≥
1
(
−
1
)
(
k
+
1
)
x
2
k
2
k
!
.
{\displaystyle 1-\cos(x)=1-\sum _{k\geq 0}{\frac {{(-1)^{(k+1)}}x^{2k}}{2k!}}=\sum _{k\geq 1}{\frac {{(-1)^{(k+1)}}x^{2k}}{2k!}}.}
Therefore,
|
1
−
cos
(
x
)
x
2
|
=
|
−
∑
k
≥
0
x
2
k
2
(
k
+
1
)
!
|
≤
∑
k
≥
0
|
x
|
k
k
!
=
e
|
x
|
.
{\displaystyle \left|{\frac {1-\cos(x)}{x^{2}}}\right|=\left|-\sum _{k\geq 0}{\frac {x^{2k}}{2(k+1)!}}\right|\leq \sum _{k\geq 0}{\frac {|x|^{k}}{k!}}=e^{|x|}.}
Splitting the integral into pieces, we have
∫
−
∞
∞
|
1
−
cos
(
x
)
x
2
|
d
x
≤
∫
−
∞
−
ε
2
x
2
d
x
+
∫
−
ε
ε
e
|
x
|
d
x
+
∫
ε
∞
2
x
2
d
x
≤
K
,
{\displaystyle \int _{-\infty }^{\infty }\left|{\frac {1-\cos(x)}{x^{2}}}\right|dx\leq \int _{-\infty }^{-\varepsilon }{\frac {2}{x^{2}}}dx+\int _{-\varepsilon }^{\varepsilon }e^{|x|}dx+\int _{\varepsilon }^{\infty }{\frac {2}{x^{2}}}dx\leq K,}
for some constant
K
>
0.
{\displaystyle K>0.}
This shows that the integral is absolutely integrable, which implies the original integral exists, and switching from
λ
{\displaystyle \lambda }
to
n
{\displaystyle n}
was in fact justified, and the proof is complete.