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Advanced Calculus for Math Enthusiasts

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asway933
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29 views3 pages

Advanced Calculus for Math Enthusiasts

Uploaded by

asway933
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Question:

Let f (x) be a real-valued function such that:

dn
[ n f (x)] = sin(n)
dx
​ ​

x=0

Evaluate:

π csc 1
∫ f (x)dx
​

(Proposed by Miracle Invoker ඞ)

Solution:
Using Maclaurin Series:
∞
xn
g(x) = ∑ g ​
(n)
(0) ​

n=0
n!

Therefore,

∞
xn
f (x) = ∑ sin(n) ​ ​

n=0
n!

Using Euler’s Formula for sin(x),


ein − e−in
sin(n) =
2i
​

∞ n ∞ n
1 ( e i x) 1 (e−i x)
f (x) = ∑ − ∑
2i 2i
​ ​ ​ ​ ​

n! n!
n=0 n=0

Recalling the Infinite Series for ex :

∞
xn
e =∑ x
​ ​

n!
n=0

i −i
ee x − ee x
f (x) =
2i
​

Using Euler’s Formula:

eix = cos x + i sin x

ei = cos 1 + i sin 1

e−i = cos 1 − i sin 1


Now,

e(cos 1+i sin 1)x − e(cos 1−i sin 1)x


f (x) =
2i
​

ex cos 1 eix sin 1 − ex cos 1 e−ix sin 1


f (x) =
2i

eix sin 1 − e−ix sin 1


f (x) = ex cos 1 ( )
2i
​

Using Euler’s Formula for sin(x):

f (x) = sin (x sin 1) ex cos 1

Applying Integration By Parts Twice results in:

∫ ex cos 1 sin (x sin 1) dx = −ex cos 1 sin (1 − x sin 1) + C

Hence,

π csc 1
∫ f (x)dx = sin (1) (1 + eπ cot 1 )
​ ​

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