Unusual Laplace Transform

Calculus Level 5

Let g ( s ) g(s) be the Laplace Transform of the function f ( x ) = 1 ⌈ x ⌉ f(x) = \dfrac{1}{\lceil x \rceil} .

What is the closed form of g g using the variable s s ?

A : ( 1 − e s ) ln ⁡ ( 1 − e − s ) s B : ln ⁡ ( e s − 1 ) s 2 C : s ln ⁡ ( e s − 1 ) e s − 1 D : ln ⁡ ( 1 − e − s ) s ( e − s − 1 ) \begin{array}{ccc} A: & \dfrac{(1-e^s)\ln(1-e^{-s})}{s} \\ B : & \dfrac{\ln(e^s-1)}{s^2} \\ C : & \dfrac{s \ln(e^s-1)}{e^s-1} \\ D : & \dfrac{\ln(1-e^{-s})}{s(e^{-s}-1)} \end{array}

Notation: ⌈ ⋅ ⌉ \lceil \cdot \rceil denotes the ceiling function .

(C) None of the given answers (A) (D) (B)

Ariel Gershon by Ariel Gershon , 26, Canada

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1 solution

Guilherme Niedu
Mar 20, 2017

g ( s ) = ∫ 0 ∞ 1 ⌈ x ⌉ e − s x d x \large \displaystyle g(s) = \int_0^{\infty} \frac{1}{\lceil x \rceil} e^{-sx} dx

g ( s ) = ∫ 0 1 1 1 e − s x d x + ∫ 1 2 1 2 e − s x d x + ∫ 2 3 1 3 e − s x d x + . . . \large \displaystyle g(s) = \int_0^1 \frac11 e^{-sx} dx + \int_1^2 \frac12 e^{-sx} dx + \int_2^3 \frac13 e^{-sx} dx + ...

g ( s ) = 1 1 e − s x − s ∣ 0 1 + 1 2 e − s x − s ∣ 1 2 + 1 3 e − s x − s ∣ 2 3 + . . . \large \displaystyle g(s) = \frac11 \frac{e^{-sx}}{-s} \Big | _0^1 + \frac12 \frac{e^{-sx}}{-s} \Big | _1^2 + \frac13 \frac{e^{-sx}}{-s} \Big | _2^3 + ...

g ( s ) = ( 1 − e − s s ) + 1 2 ( e − s − e − 2 s s ) + 1 3 ( e − 2 s − e − 3 s s ) + . . . \large \displaystyle g(s) = \Big ( \frac{1 - e^{-s}}{s} \Big ) + \frac12 \Big ( \frac{e^{-s} - e^{-2s}}{s} \Big ) + \frac13 \Big ( \frac{e^{-2s} - e^{-3s}}{s} \Big ) + ...

g ( s ) = ( 1 − e − s s ) ⋅ ( 1 + e − s 2 + e − 2 s 3 + . . . ) \large \displaystyle g(s) = \Big ( \frac{1 - e^{-s}}{s} \Big ) \cdot \Big (1 + \frac{e^{-s}}{2} + \frac{e^{-2s}}{3} + ... \Big ) \

g ( s ) = ( 1 − e − s s ) ⋅ e s ⋅ ( 1 + e − s 2 + e − 2 s 3 + . . . ) ⋅ e − s \large \displaystyle g(s) = \Big ( \frac{1 - e^{-s}}{s} \Big ) \color{#3D99F6} \cdot e^s \color{#333333} \cdot \Big (1 + \frac{e^{-s}}{2} + \frac{e^{-2s}}{3} + ... \Big ) \color{#3D99F6} \cdot e^{-s}

g ( s ) = ( e s − 1 s ) ⋅ ( e − s + e − 2 s 2 + e − 3 s 3 + . . . ) \large \displaystyle g(s) = \Big ( \frac{e^s - 1}{s} \Big ) \cdot \Big ( e^{-s} + \frac{e^{-2s}}{2} + \frac{e^{-3s}}{3} + ... \Big )

g ( s ) = ( e s − 1 s ) ⋅ ( − ln ⁡ ( 1 − e − s ) ) \large \displaystyle g(s) = \Big ( \frac{e^s - 1}{s} \Big ) \cdot (-\ln(1 - e^{-s}))

g ( s ) = ( 1 − e s ) ⋅ ln ⁡ ( 1 − e − s ) s \large \displaystyle \color{#3D99F6} g(s) = \boxed{\large \displaystyle \frac{(1-e^s)\cdot \ln(1 - e^{-s})} {s} }

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Really good Laplace Transform prob, Ariel!

tom engelsman - 4 years, 2 months ago

Perfect solution!

Thank you Tom Engelsman :)

Ariel Gershon - 4 years, 2 months ago

Log in to reply

Thanks, Ariel!

Guilherme Niedu - 4 years, 2 months ago

Nice method. Did the same way.

Indraneel Mukhopadhyaya - 4 years, 1 month ago

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