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Question Number 96198 by mathmax by abdo last updated on 30/May/20

calculate f (a) = \int_{0}^{\infty} \frac{\cos (sh (2 x))}{x^{2} + a^{2}} dx and g (a) = \int_{0}^{\infty} \frac{\cos (sh (2 x))}{{(x^{2} + a^{2})}^{2}} (a > 0)

Answered by mathmax by abdo last updated on 01/Jun/20

f (a) = \int_{0}^{\infty} \frac{\cos (sh (2 x))}{x^{2} + a^{2}} dx \Rightarrow f (a) =_{x = at} \int_{0}^{\infty} \frac{\cos (sh (2 at))}{a^{2} (t^{2} + 1)} adt

= \frac{1}{a} \int_{0}^{\infty} \frac{\cos (sh (2 at))}{t^{2} + 1} dt \Rightarrow 2 f (a) = \frac{1}{a} \int_{- \infty}^{+ \infty} \frac{\cos (sh (2 at))}{t^{2} + 1} dt

\Rightarrow 2 af (a) = Re (\int_{- \infty}^{+ \infty} \frac{e^{ish (2 at)}}{t^{2} + 1} dt) let φ (z) = \frac{e^{ish (2 az)}}{z^{2} + 1} \Rightarrow

φ (z) = \frac{e^{ish (2 az)}}{(z - i) (z + i)} residus theorem give

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, i) = 2 i π \times \frac{e^{ish (2 ai)}}{2 i} = π e^{ish (2 ai)} but

sh (2 ai) = \sin (2 a) \Rightarrow e^{ish (2 ai)} = e^{isin (2 a)} = \cos (\sin (2 a)) + isin (\sin (2 a)) \Rightarrow

2 af (a) = Re (\int_{0}^{\infty} φ (z) dz) = \cos (\sin (2 a)) \Rightarrow f (a) = \frac{\cos (\sin (2 a))}{2 a}

Commented by mathmax by abdo last updated on 01/Jun/20

we have f^{^{'}} (a) = - \int_{0}^{\infty} \frac{2 a coz (sh (2 x))}{{(x^{2} + a^{2})}^{2}} dx \Rightarrow

\int_{0}^{\infty} \frac{\cos (sh (2 x))}{{(x^{2} + a^{2})}^{2}} dx = - \frac{f^{^{'}} (a)}{2 a} we have f (a) = \frac{\cos (\sin (2 a))}{2 a} \Rightarrow

f^{^{'}} (a) = \frac{- 2 \cos (2 a) \sin (\sin (2 a) (2 a) - 2 \cos (\sin (2 a)}{{4 a}^{2}} \Rightarrow

\int_{0}^{\infty} \frac{\cos (sh (2 x))}{{(x^{2} + a^{2})}^{2}} dx = \frac{2 acos (2 a) \sin (\sin (2 a) + \cos (\sin (2 a))}{{4 a}^{3}}