L(e−at−e−btt)=∫s∞L(e−at−e−bt){s} ds=∫s∞(1s+a−1s+b) ds=ln(s+as+b)∣s∞=lims→∞ln(1+as1+bs)−ln(s+as+b)=ln(1)−ln(s+as+b)=−ln(s+as+b)=ln(s+bs+a)\begin{aligned} \mathcal{L}\left(\frac{e^{-at} - e^{-bt}}{t}\right) &= \int_s^\infty \mathcal{L}\left(e^{-at} - e^{-bt}\right)\{s\}\,\mathrm{d}s \\&= \int_s^\infty \left(\frac{1}{s + a} - \frac{1}{s + b}\right)\,\mathrm{d}s \\&= \ln\left(\frac{s + a}{s + b}\right) \vert_s^\infty \\&= \lim_{s \to \infty} \ln\left(\frac{1 + \frac{a}{s}}{1 + \frac{b}{s}}\right) - \ln\left(\frac{s + a}{s + b}\right) \\&= \ln(1) - \ln\left(\frac{s + a}{s + b}\right) = -\ln\left(\frac{s + a}{s + b}\right) \\&= \ln\left(\frac{s + b}{s + a}\right) \end{aligned}L(te−at−e−bt)=∫s∞L(e−at−e−bt){s}ds=∫s∞(s+a1−s+b1)ds=ln(s+bs+a)∣s∞=s→∞limln(1+sb1+sa)−ln(s+bs+a)=ln(1)−ln(s+bs+a)=−ln(s+bs+a)=ln(s+as+b)
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