calculate-f-0-e-x-cos-pi-x-dx-with-gt-0-

Question Number 37898 by abdo mathsup 649 cc last updated on 19/Jun/18

c a l c u l a t e f (λ) = \int_{0}^{+ \infty} e^{- λ x} c o s (π [x]) d x

w i t h λ > 0

Commented by prof Abdo imad last updated on 19/Jun/18

f (λ) = \sum_{n = 0}^{\infty} \int_{n}^{n + 1} e^{- λ x} c o s (n π) d x

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{n}^{n + 1} e^{- λ x} d x

= \sum_{n = 0}^{\infty} {(- 1)}^{n} {[- \frac{1}{λ} e^{- λ x}]}_{n}^{n + 1}

= \frac{1}{λ} \sum_{n = 0}^{\infty} {(- 1)}^{n + 1} (e^{- λ (n + 1)} - e^{- λ n})

= \frac{1}{λ} \sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- λ n} - \frac{1}{λ} \sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- λ (n + 1)}

= \frac{1}{λ} \sum_{n = 0}^{\infty} {(- e^{- λ})}^{n} - \frac{e^{- λ}}{λ} \sum_{n = 0}^{\infty} {(- e^{- λ})}^{n}

= \frac{1 - e^{- λ}}{λ} \frac{1}{1 + e^{- λ}} \Rightarrow

f (λ) = \frac{1 - e^{- λ}}{λ (1 + e^{- λ})}

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

[x] = 0 1 > x ⩾ 0

[x] = 1 2 > x ⩾ 1

[x] = 2 3 > x ⩾ 2

t h u s t h e v a l u e o f Π [x] i s i n t r e g a l m u l t i p l e o f

Π \dots h e n c e v a l u e o f c o s (Π [x]) o s c i l l a t e s

e i t t h e r + 1 o r - 1

s o t h e g i v e n i n t r e g a l h a s t w o v a l u e

1) \int_{0}^{+ \infty} e^{- λ x} \times 1 \times d x

=∣ \frac{e^{- λ x}}{- λ} ∣_{0}^{\infty}

= \frac{- 1}{λ} (\frac{1}{e^{\infty}} - \frac{1}{e^{0}}) = \frac{1}{λ}

2) \int_{0}^{+ \infty} e^{- λ x} \times - 1 \times d x

= - 1 \times ∣ \frac{e^{- λ x}}{- λ} ∣_{0}^{\infty} = \frac{1}{λ} (\frac{1}{e^{\infty}} - \frac{1}{e^{0}}) = - \frac{1}{λ}