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Jozsef Wildt International Math Competition
2019 Jozsef Wildt International Math Competition
W. 29
Prove this expression
Prove this expression
Source: 2019 Jozsef Wildt International Math Competition-W. 29
May 18, 2020
integration
Summation
Problem Statement
Prove that
∫
0
∞
e
3
t
4
e
4
t
(
3
t
−
1
)
+
2
e
2
t
(
15
t
−
17
)
+
18
(
1
−
t
)
(
1
+
e
4
t
−
e
2
t
)
2
=
12
∑
k
=
0
∞
(
−
1
)
k
(
2
k
+
1
)
2
−
10
\int \limits_0^{\infty} e^{3t}\frac{4e^{4t}(3t - 1) + 2e^{2t}(15t - 17) + 18(1 - t)}{\left(1 + e^{4t} - e^{2t}\right)^2}=12\sum \limits_{k=0}^{\infty}\frac{(-1)^k}{(2k + 1)^2}-10
0
∫
∞
e
3
t
(
1
+
e
4
t
−
e
2
t
)
2
4
e
4
t
(
3
t
−
1
)
+
2
e
2
t
(
15
t
−
17
)
+
18
(
1
−
t
)
=
12
k
=
0
∑
∞
(
2
k
+
1
)
2
(
−
1
)
k
−
10
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