3.Nevlastní integrál
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u→∞
arctg
2 u
2
−
π
2
32
=
(
π/2)
2
2
−
π
2
32
= π
2
8
−
π
2
32
= 3π
2
32
. . . evaluate the integral . . .
⊳⊳
⊳
⊲
⊲⊲
Nevlastní integrál
c
Robert Mařík, 2008 ×
Find
I =
Z
∞
1
arctg
x
x2 + 1
d
x
I = lim
u→∞
Z
u
1
arctg
x
x2 + 1
d
x
Z
arctg
x
x2 + 1
d
x
arctg
x = t
1
x2 + 1
d
x = dt
=
Z
t dt =
t
2
2
=
arctg
2 x
2
Z
u
1
arctg
x
x2 + 1
d
x =
"
arctg
2 x
2
#u
1
=
arctg
2 u
2
−
arctg
2 1
2
=
arctg
2 u
2
−
(
π/4)
2
2
=
arctg
2 u
2
−
π
2
32
I = lim
u→∞
arctg
2 u
2
−
π
2
32
=
(
π/2)
2
2
−
π
2
32
= π
2
8
−
π
2
32
= 3π
2
32
. . . and return to the variable
x.
⊳⊳
⊳
⊲
⊲⊲
Nevlastní integrál
c
Robert Mařík, 2008 ×
Find
I =
Z
∞
1
arctg
x
x2 + 1
d
x
I = lim
u→∞
Z
u
1
arctg
x
x2 + 1
d
x
Z
arctg
x
x2 + 1
d
x
arctg
x = t
1
x2 + 1
d
x = dt
=
Z
t dt =
t
2
2
=
arctg
2 x
2
Z
u
1
arctg
x
x2 + 1
d
x =
"
arctg
2 x
2
#u
1
=
arctg
2 u
2
−
arctg
2 1
2
=
arctg
2 u
2
−
(
π/4)
2
2
=
arctg
2 u
2
−
π
2
32
I = lim
u→∞
arctg
2 u
2
−
π
2
32
=
(
π/2)
2
2
−
π
2
32
= π
2
8
−
π
2
32
= 3π
2
32
We continue with the definite integral.
⊳⊳
⊳
⊲
⊲⊲
Nevlastní integrál
c
Robert Mařík, 2008 ×
Find
I =
Z
∞
1
arctg
x
x2 + 1
d
x
I = lim
u→∞
Z
u
1
arctg
x
x2 + 1
d
x
Z
arctg
x
x2 + 1
d
x
arctg
x = t
1
x2 + 1
d
x = dt
=
Z
t dt =
t
2
2
=
arctg
2 x
2
Z
u
1
arctg
x
x2 + 1
d
x =
"
arctg
2 x
2
#u
1
=
arctg
2 u
2
−
arctg
2 1
2
=
arctg
2 u
2
−
(
π/4)
2
2
=
arctg
2 u
2
−
π
2
32
I = lim
u→∞
arctg
2 u
2
−
π
2
32
=
(
π/2)
2
2
−
π
2
32
= π
2
8
−
π
2
32
= 3π
2
32
The antiderivative is known.
⊳⊳
⊳
⊲
⊲⊲
Nevlastní integrál
c
Robert Mařík, 2008 ×
Find
I =
Z
∞
1
arctg
x
x2 + 1
d
x
I = lim
u→∞
Z
u
1
arctg
x
x2 + 1
d
x
Z
arctg
x
x2 + 1
d
x
arctg
x = t
1
x2 + 1
d
x = dt
=
Z
t dt =
t
2
2
=
arctg
2 x
2
Z
u
1
arctg
x
x2 + 1
d
x =
"
arctg
2 x
2
#u
1
=
arctg
2 u
2
−
arctg
2 1
2
=
arctg
2 u
2
−
(
π/4)
2
2
=
arctg
2 u
2
−
π
2
32
I = lim
u→∞
arctg
2 u
2
−
π
2
32
=
(
π/2)
2
2
−
π
2
32
= π
2
8
−
π
2
32
= 3π
2
32
Newton–Leibniz formula yields the value of the integral.