Transform of the Logarithm Because has an infinite

Chapter 7, Problem 67E

(choose chapter or problem)

Transform of the Logarithm Because f(t) = Int has an infinite discontinuity at t = 0 it might be assumed that \(\mathscr{L}\{\ln t\}) does not exist; however, this is incorrect. The point of this problem to guide you through the formal steps leading to the Laplace transform of f(t) = Int, t  0

(a) Use integration by parts to show that

\(\mathscr{L}\{\ln t\}=s \mathscr{L}\{t \ln t\}-\frac{1}{s}\).

(b) If \\(\mathscr{L}\{\ln t\}=Y(s)\), use Theorem 7.4.1 with n = 1 to show that part (a) becomes

\(s \frac{d Y}{d s}+Y=-\frac{1}{s}\)

Find an explicit solution of the foregoing differential equation.

(c) Finally, the integral definition of Euler’s constant (sometimes called the Euler-Mascheroni constant) is \(\gamma=-\int_{0}^{\infty} e^{-t} \ln t d t\), where \(\gamma=0.5772156649\) …. Use \(Y(1)=-\gamma\) in the solution in part (b) to show that

\(\mathscr{L}\{\ln t\}=-\frac{\gamma}{s}-\frac{\ln s}{s}\), s  0

Text Transcription:

mathscr L ln t

mathscr L ln t=s mathscr L t ln t-1/s

mathscr L ln t=Y(s)

s dY/ds+Y=-1/s

gamma=-int_0^infty e^-t ln t dt

gamma=0.5772156649

Y(1)=-gamma

mathscr L ln t=-gamma/s-ln s/s