Integrating int_0^(+oo) e^(-st)cos(at)dt using the complex exponential

ormaybesaladqh

ormaybesaladqh

Answered question

2022-10-12

Integrating 0 + e s t cos ( a t ) d t using the complex exponential
Other similar questions I have found used integration by part of alternative complicated transformations that I do not understand. Using the cos ( x ) = 1 2 ( e i x + e i x ) identity, is there a way to find the solution, which should be a / ( a 2 + s 2 )?

Answer & Explanation

Milton Hampton

Milton Hampton

Beginner2022-10-13Added 16 answers

For all complex number z,
cos ( z ) = e i z + e i z 2 .
It follows that:
  0 + e s t cos ( a t ) d t = 1 2 0 + [ e s t e i a t + e s t e i a t ] d t = 1 2 0 + e ( s i a ) t d t + 1 2 0 + e ( s + i a ) t d t .
Now, if z is a non-zero complex number,
(1) 0 + e z t d t = [ 1 z e z t ] t = 0 t = + = 1 z .
Therefore, with z = s i a:
0 + e s t cos ( a t ) d t = 1 2 z + 1 2 z ¯ = z + z ¯ 2 z z ¯ = ( z ) | z | 2 = s s 2 + a 2 .
Note that (1) is true if s>0 because | e z t | = e ( z ) t = e s t 0 as t +
Nathalie Case

Nathalie Case

Beginner2022-10-14Added 4 answers

0 + e s t cos ( a t ) d t = [ e s t sin ( a t ) a ] t = 0 t = + + s a 0 + e s t sin ( a t ) d t = [ s a 2 e s t cos ( s t ) ] t = 0 t = + s 2 a 2 0 + e s t cos ( a t ) d t = s a 2 s 2 a 2 0 + e s t cos ( a t ) d t
and therefore
0 + e s t cos ( a t ) d t = s a 2 1 + s 2 a 2 = s s 2 + a 2 .

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