2. However, the Improper Riemann Integral of a function may exist without sin r the function being integrable. To see this, consider 1. a) Using integration by parts, prove that for c> , sin r -de cose 1 1. Cㅠ [ b) Deduce that the improper integral c) We will prove that [ For every nN consider (n+1) [sin r a sin r sin a dx= COS 22 dr. (k+1)x sin r Σsizdr. dr. de converges.

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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2. However, the Improper Riemann Integral of a function may exist without
sin r
the function being integrable. To see this, consider
a) Using integration by parts, prove that for c> #,
-1.²
sin r
-da
b) Deduce that the improper integral
c) We will prove that
T
For every nN consider
[+]
cosc 1
С
sin r
sin r
-dx =
sin a
-dr =
Σ
I
ESAT
COS X
22
-dz.
dr converges.
(k+1)* \sin x
I
dr.
Transcribed Image Text:2. However, the Improper Riemann Integral of a function may exist without sin r the function being integrable. To see this, consider a) Using integration by parts, prove that for c> #, -1.² sin r -da b) Deduce that the improper integral c) We will prove that T For every nN consider [+] cosc 1 С sin r sin r -dx = sin a -dr = Σ I ESAT COS X 22 -dz. dr converges. (k+1)* \sin x I dr.
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