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Question Number 107591 by hgrocks last updated on 11/Aug/20

Commented by hgrocks last updated on 11/Aug/20

$$\mathrm{Pls}\mathrm{Tell}\mathrm{Me}\mathrm{Which}\mathrm{one}\mathrm{is}\mathrm{correct}$$$$\mathrm{and}\mathrm{why}$$

Commented by hgrocks last updated on 11/Aug/20

$$\mathrm{Anyone}?$$

Answered by Aziztisffola last updated on 11/Aug/20

$$\mathrm{put}\mathrm{x}=\alpha \mathrm{t}\Rightarrow \mathrm{dx}=\alpha \mathrm{dt}\iff \mathrm{dt}=\frac{\mathrm{dx}}{\alpha }$$$$\mathrm{t}=0\Rightarrow \mathrm{x}=0\mathrm{and}\mathrm{t}\to \mathrm{\infty }\Rightarrow \mathrm{x}\to \mathrm{\infty }$$$${\int }_0^{\mathrm{\infty }}\mathrm{f}\left(\alpha \mathrm{t}\right)\mathrm{dt}=\underset{x\to \mathrm{\infty }}{\lim }{\int }_0^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)\frac{\mathrm{dx}}{\alpha }$$$$=\frac{1}{\alpha }\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }{\int }_0^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$$$$\mathrm{then}\left(1\right)\mathrm{is}\mathrm{correct}$$

Answered by mathmax by abdo last updated on 11/Aug/20

$${\int }_0^{\mathrm{\infty }}\mathrm{f}\left(\alpha \mathrm{t}\right)\mathrm{dt}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\int }_0^{\mathrm{n}}\mathrm{f}\left(\alpha \mathrm{t}\right)\mathrm{dt}{=}_{\alpha \mathrm{t}=\mathrm{x}}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\int }_0^{\mathrm{n}\alpha }\mathrm{f}\left(\mathrm{x}\right)\frac{\mathrm{dx}}{\alpha }$$$$=\frac{1}{\alpha }{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\int }_0^{\mathrm{n}\alpha }\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}.$$

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