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Albahari Integrals
Integrals (Euler's Formula)
I
1
=
∫
0
∞
e
−
s
t
cos
(
a
t
)
d
t
=
ℜ
(
I
3
)
I
2
=
∫
0
∞
e
−
s
t
sin
(
a
t
)
d
t
=
ℑ
(
I
3
)
I
3
=
I
1
+
i
I
2
=
∫
0
∞
e
−
(
s
−
i
a
)
t
d
t
=
s
s
2
+
a
2
+
i
a
s
2
+
a
2
\begin{aligned} I_1 &=\int_0^\infty e^{-st}\cos(at)dt=\Re(I_3) \\ I_2 &=\int_0^\infty e^{-st}\sin(at)dt=\Im(I_3) \\ I_3 &= I_1+i I_2=\int_0^\infty e^{-(s-ia)t}dt=\frac{s}{s^2+a^2}+i\frac{a}{s^2+a^2} \end{aligned}
I
1
I
2
I
3
=
∫
0
∞
e
−
s
t
cos
(
a
t
)
d
t
=
ℜ
(
I
3
)
=
∫
0
∞
e
−
s
t
sin
(
a
t
)
d
t
=
ℑ
(
I
3
)
=
I
1
+
i
I
2
=
∫
0
∞
e
−
(
s
−
ia
)
t
d
t
=
s
2
+
a
2
s
+
i
s
2
+
a
2
a
J
1
=
∫
0
1
e
−
t
x
cos
(
2
π
n
x
)
d
x
=
ℜ
(
J
3
)
J
2
=
∫
0
1
e
−
t
x
sin
(
2
π
n
x
)
d
x
=
ℑ
(
J
3
)
J
3
=
∫
0
1
e
−
x
(
t
−
2
π
i
n
)
d
x
=
t
(
1
−
e
−
t
)
t
2
+
(
2
π
n
)
2
+
i
2
π
n
(
1
−
e
−
t
)
t
2
+
(
2
π
n
)
2
\begin{aligned} J_1 &=\int_0^1 e^{-tx} \cos(2\pi nx)dx=\Re(J_3) \\ J_2 &=\int_0^1 e^{-tx} \sin(2\pi nx)dx=\Im(J_3) \\ J_3 &=\int_0^1 e^{-x(t-2\pi in)} dx=\frac{t(1-e^{-t})}{t^2+(2\pi n)^2} +i\frac{2\pi n(1-e^{-t})}{t^2+(2\pi n)^2} \end{aligned}
J
1
J
2
J
3
=
∫
0
1
e
−
t
x
cos
(
2
πn
x
)
d
x
=
ℜ
(
J
3
)
=
∫
0
1
e
−
t
x
sin
(
2
πn
x
)
d
x
=
ℑ
(
J
3
)
=
∫
0
1
e
−
x
(
t
−
2
πin
)
d
x
=
t
2
+
(
2
πn
)
2
t
(
1
−
e
−
t
)
+
i
t
2
+
(
2
πn
)
2
2
πn
(
1
−
e
−
t
)
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Miguel Bustamante. Last modified: August 08, 2023. Website built with
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