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Question Number 79325 by Henri Boucatchou last updated on 24/Jan/20

$$\mathit{C}\mathit{o}\mathit{n}\mathit{v}\mathit{e}\mathit{r}\mathit{g}\mathit{e}\mathit{n}\mathit{c}\mathit{e}\mathit{o}\mathit{f}:$$$$1)\mathit{I}={\int }_1^{\mathrm{\infty }}\frac{{\mathit{e}}^{-\mathit{t}/5}\mid \mathit{s}\mathit{i}\mathit{n}\left(\mathit{l}\mathit{n}\mathit{t}\right)\mid }{{(\mathit{t}-1)}^{3/2}}\mathit{d}\mathit{t}$$$$2)\mathit{I}={\int }_1^{\mathrm{\infty }}\frac{\sqrt{\mathit{l}\mathit{n}\mathit{x}}}{(\mathit{x}-1)\sqrt{\mathit{x}}}\mathit{d}\mathit{x}$$

Commented by mathmax by abdo last updated on 24/Jan/20

$$1)changementt-1=xgive$$$$I={\int }_1^{+\mathrm{\infty }}\frac{e^{-\frac{t}{5}}\mid sin\left(lnt\right)\mid }{{(t-1)}^{\frac{3}{2}}}dt={\int }_0^{\mathrm{\infty }}\frac{e^{-\frac{1+x}{5}}\mid sin(ln(1+x)\mid }{x^{\frac{3}{2}}}dx={\int }_0^{\mathrm{\infty }}\phi \left(x\right)dx$$$$I={\int }_0^1(...)dx+{\int }_1^{+\mathrm{\infty }}(...)dxatV\left(0\right)\mid sin(ln(1+x)\mid \sim \mid sin\left(x\right)\mid \sim \mid x\mid$$$$\Rightarrow \phi \left(x\right)\sim \frac{e^{-\frac{1}{5}}x}{x^{\frac{3}{2}}}=\frac{e^{-\frac{1}{5}}}{x^{\frac{1}{2}}}and{\int }_0^1{e}^{-\frac{1}{5}}{x}^{-\frac{1}{2}}dxconverges$$$$at+\mathrm{\infty }\mid \phi \left(x\right)\mid ⩽\frac{e^{-\frac{1+x}{5}}}{x^{\frac{3}{2}}}\Rightarrow {\int }_1^{+\mathrm{\infty }}\mid \phi \left(x\right)\mid dx⩽{\int }_1^{+\mathrm{\infty }}{x}^{-\frac{3}{2}}{e}^{-\frac{1+x}{5}}dx$$$$wehavelim{x}^2{x}^{-\frac{3}{2}}{e}^{-\frac{1+x}{5}}=0\Rightarrow thisintegralconverges\Rightarrow$$$$Iconverges$$

Commented by mathmax by abdo last updated on 24/Jan/20

$$2)I={\int }_1^{+\mathrm{\infty }}\frac{\sqrt{lnx}}{(x-1)\sqrt{x}}dxchangement\sqrt{lnx}=tgivelnx=t^2\Rightarrow$$$$x=e^{t^2}\Rightarrow I={\int }_0^{\mathrm{\infty }}\frac{t}{(e^{t^2}-1)e^{\frac{t^2}{2}}}\times 2t{e}^{t^2}dt={\int }_0^{\mathrm{\infty }}\frac{2t^2{e}^{\frac{t^2}{2}}}{e^{t^2}-1}dt$$$${=}_{}{\int }_0^{\mathrm{\infty }}\frac{2t^2{e}^{-t^2}{e}^{\frac{t^2}{2}}}{1-e^{-t^2}}dt={\int }_0^{\mathrm{\infty }}\frac{2t^2{e}^{-\frac{t^2}{2}}}{1-e^{-t^2}}dt={\int }_0^1(...)dt+{\int }_1^{+\mathrm{\infty }}(...)dt$$$$atV\left(0\right)e^{-t^2}\sim 1-t^2\Rightarrow 1-e^{-t^2}\sim 1-1+t^2\Rightarrow \Rightarrow \frac{2t^2{e}^{-\frac{t^2}{2}}}{1-e^{-t^2}}\sim 2e^{-\frac{t^2}{2}}\to 2(t\to 0)$$$$\Rightarrow {\int }_0^1\frac{2t{e}^{-\frac{t^2}{2}}}{1-e^{-t^2}}dtconvergeslet\xi >0wehave$$$${lim}_{t\to +\mathrm{\infty }}{t}^{1+\xi }\times \frac{2t{e}^{-\frac{t^2}{2}}}{1-e^{-t^2}}=0\Rightarrow {\int }_1^{+\mathrm{\infty }}\frac{2t^2{e}^{-\frac{t^2}{2}}}{1-e^{-t^2}}converges$$$$finallyIisconvergent$$

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