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Putnam
1968 Putnam
B4
Putnam 1968 B4
Putnam 1968 B4
Source: Putnam 1968
February 19, 2022
Putnam
integration
Measure theory
Problem Statement
Suppose that
f
:
R
→
R
f:\mathbb{R} \rightarrow \mathbb{R}
f
:
R
→
R
is continuous and
L
=
∫
−
∞
∞
f
(
x
)
d
x
L=\int_{-\infty}^{\infty} f(x) dx
L
=
∫
−
∞
∞
f
(
x
)
d
x
exists. Show that
∫
−
∞
∞
f
(
x
−
1
x
)
d
x
=
L
.
\int_{-\infty}^{\infty}f\left(x-\frac{1}{x}\right)dx=L.
∫
−
∞
∞
f
(
x
−
x
1
)
d
x
=
L
.
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