Question Number 177493 by ali009 last updated on 06/Oct/22
$${find}\:{the}\:{laplace}\:{transform}\:{of} \\ $$$${f}\left({t}\right)={ln}\left({t}\right) \\ $$
Commented by JDamian last updated on 06/Oct/22
Is it suitable for Laplace Transform?
Commented by ali009 last updated on 06/Oct/22
$${yes}\:{i}\:{will}\:{put}\:{my}\:{solution}\:{later}\: \\ $$
Answered by ali009 last updated on 06/Oct/22
![in general L[t^v ln(t)]=∫_0 ^∞ t^v ln(t)e^(−st) dt let x=st dx=sdt L[t^v ln(t)]=(1/s)∫_0 ^∞ e^(−x) ((x/s))^v ln((x/s))dx =(1/s^(v+1) )∫_0 ^∞ e^(−x) x^v (ln(x)−ln(s))dx =(1/s^(v+1) )(∫_0 ^∞ e^(−x) ln(x) x^v dx −ln(s)∫_0 ^∞ e^(−x) x^v ) =(1/s^(v+1) )((d/dv)∫e^(−x) x^v dx − ln(s)Γ(v+1)) =(1/s^(v+1) )((d/dv)Γ(v+1)−ln(s)Γ(v+1)) =(1/s^(v+1) )Γ(v+1)(((Γ′(v+1))/(Γ(v+1)))−ln(s)) =(1/s^(v+1) )Γ(v+1)(ψ(v+1)−ln(s)) when t=0 L[ln(t)]=(1/s^(v+1) )Γ(1)(ψ(1)−ln(s)) L[ln(t)]=((−γ−ln(s))/s) γ is Euler′s constant](https://www.tinkutara.com/question/Q177543.png)
$${in}\:{general}\:\mathcal{L}\left[{t}^{{v}} {ln}\left({t}\right)\right]=\int_{\mathrm{0}} ^{\infty} {t}^{{v}} {ln}\left({t}\right){e}^{−{st}} {dt} \\ $$$${let}\:{x}={st}\:{dx}={sdt} \\ $$$$\mathcal{L}\left[{t}^{{v}} {ln}\left({t}\right)\right]=\frac{\mathrm{1}}{{s}}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} \left(\frac{{x}}{{s}}\right)^{{v}} {ln}\left(\frac{{x}}{{s}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{{s}^{{v}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{v}} \left({ln}\left({x}\right)−{ln}\left({s}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{{s}^{{v}+\mathrm{1}} }\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {ln}\left({x}\right)\:{x}^{{v}} \:{dx}\:−{ln}\left({s}\right)\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{v}} \right) \\ $$$$=\frac{\mathrm{1}}{{s}^{{v}+\mathrm{1}} }\left(\frac{{d}}{{dv}}\int{e}^{−{x}} {x}^{{v}} \:{dx}\:−\:{ln}\left({s}\right)\Gamma\left({v}+\mathrm{1}\right)\right) \\ $$$$=\frac{\mathrm{1}}{{s}^{{v}+\mathrm{1}} }\left(\frac{{d}}{{dv}}\Gamma\left({v}+\mathrm{1}\right)−{ln}\left({s}\right)\Gamma\left({v}+\mathrm{1}\right)\right) \\ $$$$=\frac{\mathrm{1}}{{s}^{{v}+\mathrm{1}} }\Gamma\left({v}+\mathrm{1}\right)\left(\frac{\Gamma'\left({v}+\mathrm{1}\right)}{\Gamma\left({v}+\mathrm{1}\right)}−{ln}\left({s}\right)\right) \\ $$$$=\frac{\mathrm{1}}{{s}^{{v}+\mathrm{1}} }\Gamma\left({v}+\mathrm{1}\right)\left(\psi\left({v}+\mathrm{1}\right)−{ln}\left({s}\right)\right) \\ $$$${when}\:{t}=\mathrm{0} \\ $$$$\mathcal{L}\left[{ln}\left({t}\right)\right]=\frac{\mathrm{1}}{{s}^{{v}+\mathrm{1}} }\Gamma\left(\mathrm{1}\right)\left(\psi\left(\mathrm{1}\right)−{ln}\left({s}\right)\right) \\ $$$$\mathcal{L}\left[{ln}\left({t}\right)\right]=\frac{−\gamma−{ln}\left({s}\right)}{{s}} \\ $$$$\gamma\:{is}\:{Euler}'{s}\:{constant} \\ $$