3.Nevlastní integrál
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Z
e
x
1 + (
ex)2
d
x
ex = t
ex dx = dt
=
Z
1
1 +
t2
d
t = arctg t
= arctg ex
Z
0
u
1
e−x + ex
d
x = [arctg ex]
0
u
= arctg e0 − arctg eu = arctg 1 − arctg eu
= π
4
− arctg e
u
Z
0
−∞
1
e−x + ex
d
x = lim
u→−∞
π
4
− arctg e
u
= π
4
− arctg e−∞ =
π
4
− arctg 0
=
π
4
Z
1
e−x + ex
d
x = arctg ex
Z
0
−∞
1
e−x + ex
d
x =
π
4
Z
u
1
x u
u
0
u
. . . and substitute.
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Nevlastní integrál
c
Robert Mařík, 2008 ×
Find
I =
Z
∞
−∞
1
e−x + ex
d
x
Z
∞
−∞
1
e−x + ex
d
x =
Z
0
−∞
1
e−x + ex
d
x +
Z
∞
0
1
e−x + ex
d
x
= lim
u→−∞
Z
0
u
1
e−x + ex
d
x + lim
u→∞
Z
u
0
1
e−x + ex
d
x
Z
1
e−x + ex
d
x =
Z
e
x
1 + (
ex)2
d
x
ex = t
ex dx = dt
=
Z
1
1 +
t2
d
t = arctg t
= arctg ex
Z
0
u
1
e−x + ex
d
x = [arctg ex]
0
u
= arctg e0 − arctg eu = arctg 1 − arctg eu
= π
4
− arctg e
u
Z
0
−∞
1
e−x + ex
d
x = lim
u→−∞
π
4
− arctg e
u
= π
4
− arctg e−∞ =
π
4
− arctg 0
=
π
4
Z
1
e−x + ex
d
x = arctg ex
Z
0
−∞
1
e−x + ex
d
x =
π
4
Z
u
1
x u
u
0
u
The substitution gives this integral. . .
⊳⊳
⊳
⊲
⊲⊲
Nevlastní integrál
c
Robert Mařík, 2008 ×
Find
I =
Z
∞
−∞
1
e−x + ex
d
x
Z
∞
−∞
1
e−x + ex
d
x =
Z
0
−∞
1
e−x + ex
d
x +
Z
∞
0
1
e−x + ex
d
x
= lim
u→−∞
Z
0
u
1
e−x + ex
d
x + lim
u→∞
Z
u
0
1
e−x + ex
d
x
Z
1
e−x + ex
d
x =
Z
e
x
1 + (
ex)2
d
x
ex = t
ex dx = dt
=
Z
1
1 +
t2
d
t = arctg t
= arctg ex
Z
0
u
1
e−x + ex
d
x = [arctg ex]
0
u
= arctg e0 − arctg eu = arctg 1 − arctg eu
= π
4
− arctg e
u
Z
0
−∞
1
e−x + ex
d
x = lim
u→−∞
π
4
− arctg e
u
= π
4
− arctg e−∞ =
π
4
− arctg 0
=
π
4
Z
1
e−x + ex
d
x = arctg ex
Z
0
−∞
1
e−x + ex
d
x =
π
4
Z
u
1
x u
u
0
u
. . . which can be integrated by direct formula.
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