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Question Number 200155 by Calculusboy last updated on 15/Nov/23

Answered by Frix last updated on 15/Nov/23

$$s>0$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\sqrt{t}{\mathrm{e}}^{-st}dt=$$$$=-\frac{1}{s}{\left[\sqrt{t}{\mathrm{e}}^{-st}\right]}_0^{\mathrm{\infty }}+\frac{1}{2s}\times \underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{\mathrm{e}}^{-st}}{\sqrt{t}}dt$$$$-\frac{1}{s}{\left[\sqrt{t}{\mathrm{e}}^{-st}\right]}_0^{\mathrm{\infty }}=0$$$$\frac{1}{2s}t\times \underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{\mathrm{e}}^{-st}}{\sqrt{t}}dt\stackrel{u=\sqrt{st}}{=}{s}^{-\frac{3}{2}}\times \underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{e}}^{-u^2}du=\frac{\sqrt{\pi s^{-3}}}{2}$$

Commented by Calculusboy last updated on 15/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

Answered by Mathspace last updated on 15/Nov/23

$$I={\int }_0^{\mathrm{\infty }}\sqrt{t}{e}^{-st}dt{=}_{st=x}$$$${\int }_0^{\mathrm{\infty }}\sqrt{\frac{x}{s}}{e}^{-x}\frac{dx}{s}$$$$=\frac{1}{s^{\frac{3}{2}}}{\int }_0^{\mathrm{\infty }}{x}^{\frac{1}{2}}{e}^{-x}dx$$$$=\frac{1}{s^{\frac{3}{2}}}\times \mathrm{\Gamma }\left(\frac{3}{2}\right)=\frac{1}{s^{\frac{3}{2}}}\frac{1}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)$$$$=\frac{\sqrt{\pi }}{2s^{\frac{3}{2}}}$$

Commented by Calculusboy last updated on 16/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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