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Ramprax (talk ) 11:42, 4 February 2020 (UTC) [ reply ]
D
n
(
x
)
=
∑
k
=
−
n
n
e
2
π
i
k
x
=
sin
(
(
2
n
+
1
)
π
x
)
sin
(
π
x
)
{\displaystyle D_{n}(x)=\sum _{k=-n}^{n}e^{2\pi ikx}={\frac {\sin \left(\left(2n+1\right)\pi x\right)}{\sin(\pi x)}}}
be the Dirichlet kernel .
This is clearly symmetric about zero, that is,
D
n
(
−
x
)
=
D
n
(
x
)
{\displaystyle D_{n}(-x)=D_{n}(x)}
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 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}
b
{\displaystyle b}
a
<
0
<
b
{\displaystyle a<0<b}
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}}}
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.}
r
→
⋅
n
^
=
d
=
a
→
⋅
n
^
.
{\displaystyle {\vec {r}}\cdot {\hat {n}}=d={\vec {a}}\cdot {\hat {n}}.}
∫
e
k
x
sin
(
n
x
)
d
x
=
e
k
x
k
2
+
n
2
(
k
sin
(
n
x
)
−
n
cos
(
n
x
)
)
{\displaystyle \int e^{kx}\sin(nx)\,dx={\frac {e^{kx}}{k^{2}+n^{2}}}\left(k\sin(nx)-n\cos(nx)\right)}
∫
e
k
x
cos
(
n
x
)
d
x
=
e
k
x
k
2
+
n
2
(
k
cos
(
n
x
)
+
n
sin
(
n
x
)
)
{\displaystyle \int e^{kx}\cos(nx)\,dx={\frac {e^{kx}}{k^{2}+n^{2}}}\left(k\cos(nx)+n\sin(nx)\right)}
Ramprax (talk ) 11:42, 4 February 2020 (UTC) [ reply ]
