Evaluate:
If can be expressed as , where and is square-free, then enter your answer as .
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∫ 0 ∞ 2 c o s h x − 1 e x − 1 d x = 2 ∫ 0 ∞ t 4 + t 2 + 1 t 2 d t e x = 1 + t 2 = 2 ∫ 0 ∞ u 4 + u 2 + 1 d u t → u 1 = ∫ 0 ∞ x 2 + x + 1 x − 1 / 2 d x u 2 → x = 3 π
In the last line we have used F ( a , s ) = ∫ 0 ∞ x 2 + 2 a x + 1 x s d x = sin ( π ∣ s ∣ ) 1 − a 2 π sin ( ( 1 − s ) cos − 1 a ) which I have proved here