How to Calculate the Laplace Transform of the Natural Logarithm

Опубликовал Admin
1-01-2017, 15:06
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The Laplace transform is an integral transform widely used to solve differential equations with constant coefficients. The transforms are typically very straightforward, but there are functions whose Laplace transforms cannot easily be found using elementary methods. In this article, we show how to obtain the Laplace transform of the natural logarithm using expansions of the Gamma function. Thus, it is recommended that you be familiar with the basic properties of this function before proceeding.

Steps

  1. Begin with the integral. This is an integral that involves the logarithmic function. No amount of integration by parts, u-substitution, or any other technique learned in introductory calculus class will solve this integral, because this integrand does not have an antiderivative that can be written in terms of elementary functions.
    • ∫0∞ln⁡te−stdt{\displaystyle \int _{0}^{\infty }\ln te^{-st}{\mathrm {d} }t}
  2. Make the u-sub u=st{\displaystyle u=st}. By the properties of the log, the integral is split into two. The latter is easy to evaluate using the fundamental theorem because s{\displaystyle s} is independent of u.{\displaystyle u.}
    • 1s∫0∞ln(us)e−udu=1s∫0∞ln⁡ue−udu−ln⁡ss{\displaystyle {\frac {1}{s}}\int _{0}^{\infty }\ln \left({\frac {u}{s}}\right)e^{-u}{\mathrm {d} }u={\frac {1}{s}}\int _{0}^{\infty }\ln ue^{-u}{\mathrm {d} }u-{\frac {\ln s}{s}}}
  3. Consider the series expansion of the Gamma function. There are two important formulas to consider here.
    • The first is given below. It is a formula that expresses the logarithm of the Gamma function as an infinite series. This formula is derived from the infinite product definition (see the tips), where ϵ{\displaystyle \epsilon } is a small number, γ≈0.577...{\displaystyle \gamma \approx 0.577...} is the Euler-Mascheroni constant, and ζ(j){\displaystyle \zeta (j)} is the Riemann zeta function. (Don't worry about the summation part - it turns out that it won't be important for what we're about to do.)
      • ln⁡Γ(1+ϵ)=−γϵ+∑j=2∞(−1)jζ(j)jϵj{\displaystyle \ln \Gamma (1+\epsilon )=-\gamma \epsilon +\sum _{j=2}^{\infty }{\frac {(-1)^{j}\zeta (j)}{j}}\epsilon ^{j}}
    • The second comes straight from the integral definition of the Gamma function, Legendre's expression. We rewrite the integral so as to write the exponent with e{\displaystyle e} in the base, and rewrite that in terms of its Taylor series.
      • Γ(1+ϵ)=∫0∞xϵe−xdx=∑n=0∞ϵnn!∫0∞lnn⁡xe−xdx{\displaystyle \Gamma (1+\epsilon )=\int _{0}^{\infty }x^{\epsilon }e^{-x}{\mathrm {d} }x=\sum _{n=0}^{\infty }{\frac {\epsilon ^{n}}{n!}}\int _{0}^{\infty }\ln ^{n}xe^{-x}{\mathrm {d} }x}
  4. Find the coefficient of ϵ{\displaystyle \epsilon }. Specifically, ϵ{\displaystyle \epsilon } to the first power. The reason why is because the integral we want to compute is in the coefficient of the Taylor series for the exponential function. The specific integral sets n=1,{\displaystyle n=1,} so to evaluate the integral, we need to equate the two expressions. We first look at the first formula and take the exponent of both sides.
    • Γ(1+ϵ)=exp(−γϵ+∑j=2∞(−1)jζ(j)jϵj){\displaystyle \Gamma (1+\epsilon )=\exp \left(-\gamma \epsilon +\sum _{j=2}^{\infty }{\frac {(-1)^{j}\zeta (j)}{j}}\epsilon ^{j}\right)}
    • Since ϵ{\displaystyle \epsilon } is a small number, we can safely neglect any higher-order terms. This is why we don't need to worry about the summation part.
      • Γ(1+ϵ)≈e−γϵ≈1−γϵ{\displaystyle \Gamma (1+\epsilon )\approx e^{-\gamma \epsilon }\approx 1-\gamma \epsilon }
  5. Evaluate the integral in step 2 by equating the coefficients. Combining our previous results, we have arrived at the Laplace transform of the natural logarithm.
    • ∫0∞ln⁡ue−udu=−γ{\displaystyle \int _{0}^{\infty }\ln ue^{-u}{\mathrm {d} }u=-\gamma }
    • L{ln⁡t}=−γ+ln⁡ss{\displaystyle {\mathcal {L}}\{\ln t\}=-{\frac {\gamma +\ln s}{s}}}

Tips

  • Weierstrauss's infinite product expression for the Gamma function is given below. It is from here (and the recursion relation) that we can derive ln⁡Γ(1+ϵ).{\displaystyle \ln \Gamma (1+\epsilon ).}
    • Γ(1+ϵ)=ϵΓ(ϵ)=e−γϵ∏k=1∞eϵ/k(1+ϵk)−1{\displaystyle \Gamma (1+\epsilon )=\epsilon \Gamma (\epsilon )=e^{-\gamma \epsilon }\prod _{k=1}^{\infty }e^{\epsilon /k}\left(1+{\frac {\epsilon }{k}}\right)^{-1}}
    • ln⁡Γ(1+ϵ)=−γϵ+∑k=1∞(ϵk−ln(1+ϵk))=−γϵ+∑k=1∞(ϵk−∑j=1∞(−1)j+1ϵjjkj)=−γϵ+∑k=1∞∑j=2∞(−1)jϵjjkj=−γϵ+∑j=2∞(−1)jζ(j)jϵj{\displaystyle {\begin{aligned}\ln \Gamma (1+\epsilon )&=-\gamma \epsilon +\sum _{k=1}^{\infty }\left({\frac {\epsilon }{k}}-\ln \left(1+{\frac {\epsilon }{k}}\right)\right)\\&=-\gamma \epsilon +\sum _{k=1}^{\infty }\left({\frac {\epsilon }{k}}-\sum _{j=1}^{\infty }{\frac {(-1)^{j+1}\epsilon ^{j}}{jk^{j}}}\right)\\&=-\gamma \epsilon +\sum _{k=1}^{\infty }\sum _{j=2}^{\infty }{\frac {(-1)^{j}\epsilon ^{j}}{jk^{j}}}\\&=-\gamma \epsilon +\sum _{j=2}^{\infty }{\frac {(-1)^{j}\zeta (j)}{j}}\epsilon ^{j}\end{aligned}}}
    • In the last step, we used the summation definition of the Riemann zeta function.
    • ζ(j)=∑k=1∞1kj{\displaystyle \zeta (j)=\sum _{k=1}^{\infty }{\frac {1}{k^{j}}}}
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