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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
75
75
Part of
2005 Today's Calculation Of Integral
Problems
(1)
Today's calculation of Integral 75
Source: 1970 Tokyo Institute of Technology
8/26/2005
A function
f
(
θ
)
f(\theta)
f
(
θ
)
satisfies the following conditions
(
a
)
,
(
b
)
(a),(b)
(
a
)
,
(
b
)
.
(
a
)
f
(
θ
)
≥
0
(a)\ f(\theta)\geq 0
(
a
)
f
(
θ
)
≥
0
(
b
)
∫
0
π
f
(
θ
)
sin
θ
d
θ
=
1
(b)\ \int_0^{\pi} f(\theta)\sin \theta d\theta =1
(
b
)
∫
0
π
f
(
θ
)
sin
θ
d
θ
=
1
Prove the following inequality.
∫
0
π
f
(
θ
)
sin
n
θ
d
θ
≤
n
(
n
=
1
,
2
,
⋯
)
\int_0^{\pi} f(\theta)\sin n\theta \ d\theta \leq n\ (n=1,2,\cdots)
∫
0
π
f
(
θ
)
sin
n
θ
d
θ
≤
n
(
n
=
1
,
2
,
⋯
)
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