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Question Number 77752 by abdomathmax last updated on 09/Jan/20

l e t f (λ) = \int_{- \infty}^{+ \infty} \frac{s i n (λ e^{x} + e^{- x})}{x^{2} + λ^{2}} d x w i t h λ ⩾ 0

1) d e t d r m i n e a e x p l i c i t f o r m o f f (λ)

2) c a l c u l a t e f^{^{'}} (λ) a t f o r m o f i n t e r g r a l a n d f i n d

i t s v a l u e .

Commented by mathmax by abdo last updated on 10/Jan/20

f (λ) = \int_{- \infty}^{+ \infty} \frac{s i n (λ e^{x} + e^{- x})}{x^{2} + λ^{2}} \Rightarrow f (λ) =_{x = λ t} \int_{- \infty}^{+ \infty} \frac{s i n (λ e^{λ t} + e^{- λ t})}{λ^{2} (t^{2} + 1)} λ (d t)

= \frac{1}{λ} \int_{- \infty}^{+ \infty} \frac{s i n (λ e^{λ t} + e^{- λ t})}{t^{2} + 1} d t (w e t a k e λ > 0) \Rightarrow

λ f (λ) = I m (\int_{- \infty}^{+ \infty} \frac{e^{i (λ e^{λ t} + e^{- λ t})}}{t^{2} + 1} d t) l e t W (z) = \frac{e^{i (λ e^{λ z} + e^{- λ z})}}{z^{2} + 1}

\Rightarrow W (z) = \frac{e^{i (λ e^{λ z} + e^{- λ z})}}{(z - i) (z + i)} a n d \int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i)

= 2 i π \times \frac{e^{i (λ e^{i λ} + e^{- i λ})}}{2 i} w e h a v e

i (λ e^{i λ} + e^{- i λ}) = i (λ c o s (λ) + λ i s i n (λ) + c o s (λ) - i s i n (λ))

= i λ c o s (λ) - λ s i n (λ) + i c o s (λ) + s i n λ

= (1 - λ) s i n λ + i (1 + λ) c o s (λ) \Rightarrow

e^{i (λ e^{i λ} + e^{- i λ)}} = e^{(1 - λ) s i n (λ)} {c o s (1 + λ) c o s λ + i s i n (1 + λ) c o s λ} \Rightarrow

\int_{- \infty}^{+ \infty} W (z) d z = π e^{(1 - λ) s i n λ} c o s ((1 + λ) c o s λ) + i π e^{(1 - λ) s i n λ} s i n ((1 + λ) c o s λ)

\Rightarrow λ f (λ) = π e^{(1 - λ) s i n λ} s i n {(1 + λ) c o s λ} \Rightarrow

f (λ) = \frac{π}{λ} e^{(1 - λ) s i n λ} s i n {(1 + λ) c o s λ} (λ > 0)