I would like to evaluate the integral below ∫ 0 ∞ sin 3 ⁡ x...

migongoniwt

migongoniwt

Answered

2022-06-27

I would like to evaluate the integral below
0 sin 3 x cosh x + cos x d x x

Answer & Explanation

scipionhi

scipionhi

Expert

2022-06-28Added 25 answers

Since
sin ( x ) cosh ( x ) + cos ( x ) = 2 n 1 ( 1 ) n 1 e n x sin ( n x )
we can combine the above with
sin 2 ( x ) sin ( n x ) = 1 4 [ 2 sin ( n x ) + sin ( ( 2 n ) x ) sin ( ( 2 + n ) x ) ] and 0 e a x sin ( b x ) x d x = arctan ( b a ) to get
I = 1 2 n 1 ( 1 ) n 1 0 e n x x [ 2 sin ( n x ) + sin ( ( 2 n ) x ) sin ( ( 2 + n ) x ) ] d x = 1 2 n 1 ( 1 ) n 1 [ π 2 + arctan ( 2 n n ) arctan ( 2 + n n ) ] = 1 2 n 1 ( 1 ) n 1 [ arctan ( n + 1 ) arctan ( n 1 ) ] = 1 2 [ arctan ( 0 ) + arctan ( 1 ) ] = π 8
as desired.
pokoljitef2

pokoljitef2

Expert

2022-06-29Added 9 answers

It was shown in this answer that if n is a nonnegative integer, then
0 sin ( ( 2 n + 1 ) x ) cosh ( x ) + cos ( x ) d x x = π 4 .
Therefore,
0 sin 3 ( x ) cosh ( x ) + cos ( x ) d x x = 1 4 0 3 sin ( x ) sin ( 3 x ) cosh ( x ) + cos ( x ) d x x = 1 4 ( 3 π 4 π 4 ) = π 8 .
In general, if m is a nonnegative integer, then
0 sin 2 m + 1 ( x ) cosh ( x ) + cos ( x ) d x x = ( 1 ) 1 2 2 m k = 0 m ( 1 ) m k ( 2 m + 1 k ) 0 sin ( ( 2 m + 1 2 k ) x ) cosh ( x ) + cos ( x ) d x x = π 2 2 m + 2 k = 0 m ( 1 ) m k ( 2 m + 1 k ) .

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