0-e-st-cosh-2t-cosh-5t-dt-t-

Question Number 150259 by mathdanisur last updated on 10/Aug/21

\underset{0}{\int^{\infty}} \frac{e^{- st} (\cosh (2 t) - \cosh (5 t)) dt}{t} = ?

Answered by Ar Brandon last updated on 10/Aug/21

I (s) = \int_{0}^{\infty} \frac{e^{- s t} (\cosh (2 t) - \cosh (5 t))}{t} d t

I^{'} (s) = - \int_{0}^{\infty} e^{- s t} (\cosh (2 t) - \cosh (5 t)) d t

= - \frac{1}{2} \int_{0}^{\infty} e^{- s t} (e^{2 t} + e^{- 2 t} - e^{5 t} - e^{- 5 t}) d t

= - \frac{1}{2} {[\frac{e^{- (s - 2) t}}{2 - s} - \frac{e^{- (s + 2) t}}{s + 2} - \frac{e^{- (s - 5) t}}{5 - s} + \frac{e^{- (s + 5) t}}{s + 5}]}_{0}^{\infty}

= \frac{1}{2} (\frac{1}{2 - s} - \frac{1}{s + 2} - \frac{1}{5 - s} + \frac{1}{s + 5})

I (s) = \frac{1}{2} (\ln ∣ s + 5 ∣ + \ln ∣ s - 5 ∣ - \ln ∣ s + 2 ∣ - \ln ∣ s - 2 ∣) + C

= \frac{1}{2} \ln ∣ \frac{s^{2} - 25}{s^{2} - 4} ∣ + C . I (+ \infty) = 0 = C

\Rightarrow I (s) = \frac{1}{2} \ln ∣ \frac{s^{2} - 25}{s^{2} - 4} ∣

Commented by mathdanisur last updated on 10/Aug/21

Cool S er, Thank Y ou

Commented by Ar Brandon last updated on 10/Aug/21

L^{- 1} {\frac{s + 1}{s^{2} - 16 s + 100}} = L^{- 1} {\frac{s + 1}{{(s - 8)}^{2} + 6^{2}}}

= L^{- 1} {\frac{s - 8}{{(s - 8)}^{2} + 6^{2}} + \frac{9}{{(s - 8)}^{2} + 6^{2}}}

= \cos (6 (s - 8)) + \frac{9}{6} \sin (6 (s - 8))

Commented by Ar Brandon last updated on 10/Aug/21

You deleted your previous post .

Commented by mathdanisur last updated on 10/Aug/21

Cool Ser, Thank You