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Question Number 177493 by ali009 last updated on 06/Oct/22

$$findthelaplacetransformof$$$$f\left(t\right)=ln\left(t\right)$$

Commented by JDamian last updated on 06/Oct/22

Is it suitable for Laplace Transform?

Commented by ali009 last updated on 06/Oct/22

$$yesiwillputmysolutionlater$$

Answered by ali009 last updated on 06/Oct/22

$$ingeneral\mathcal{L}\left[t^{v}ln\left(t\right)\right]={\int }_0^{\mathrm{\infty }}{t}^{v}ln\left(t\right)e^{-st}dt$$$$letx=stdx=sdt$$$$\mathcal{L}\left[t^{v}ln\left(t\right)\right]=\frac{1}{s}{\int }_0^{\mathrm{\infty }}{e}^{-x}{\left(\frac{x}{s}\right)}^{v}ln\left(\frac{x}{s}\right)dx$$$$=\frac{1}{s^{v+1}}{\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^v(ln\left(x\right)-ln\left(s\right))dx$$$$=\frac{1}{s^{v+1}}({\int }_0^{\mathrm{\infty }}{e}^{-x}ln\left(x\right)x^{v}dx-ln\left(s\right){\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^v)$$$$=\frac{1}{s^{v+1}}(\frac{d}{dv}\int e^{-x}{x}^{v}dx-ln\left(s\right)\mathrm{\Gamma }(v+1))$$$$=\frac{1}{s^{v+1}}(\frac{d}{dv}\mathrm{\Gamma }(v+1)-ln\left(s\right)\mathrm{\Gamma }(v+1))$$$$=\frac{1}{s^{v+1}}\mathrm{\Gamma }(v+1)(\frac{{\mathrm{\Gamma }}^{\prime }(v+1)}{\mathrm{\Gamma }(v+1)}-ln\left(s\right))$$$$=\frac{1}{s^{v+1}}\mathrm{\Gamma }(v+1)(\psi (v+1)-ln\left(s\right))$$$$whent=0$$$$\mathcal{L}\left[ln\left(t\right)\right]=\frac{1}{s^{v+1}}\mathrm{\Gamma }\left(1\right)(\psi \left(1\right)-ln\left(s\right))$$$$\mathcal{L}\left[ln\left(t\right)\right]=\frac{-\gamma -ln\left(s\right)}{s}$$$$\gamma is{Euler}^{\prime }sconstant$$

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