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Question Number 139878 by bramlexs22 last updated on 02/May/21

$$\mathrm{I}\left(\alpha \right)=\underset{\alpha }{\stackrel{{\alpha }^2}{\int }}\frac{\sin \alpha \mathrm{x}}{\mathrm{x}}\mathrm{dx}$$$$\frac{\mathrm{dI}\left(\alpha \right)}{\mathrm{d}\alpha }=?$$

Answered by EDWIN88 last updated on 02/May/21

$$\frac{\mathrm{dI}}{\mathrm{d}\alpha }=\underset{\alpha }{\stackrel{{\alpha }^2}{\int }}\frac{\partial }{\partial \alpha }\left(\frac{\sin \alpha \mathrm{x}}{\mathrm{x}}\right)\mathrm{dx}+\frac{\sin {\alpha }^3}{{\alpha }^2}\left(2\alpha \right)-\frac{\sin {\alpha }^2}{\alpha }\left(1\right)$$$$=\underset{\alpha }{\stackrel{{\alpha }^2}{\int }}\cos \alpha \mathrm{x}\mathrm{dx}+\frac{2\sin {\alpha }^3-\sin {\alpha }^2}{\alpha }$$$$={\left[\frac{1}{\alpha }\sin \alpha \mathrm{x}\right]}_{\alpha }^{{\alpha }^2}+\frac{2\sin {\alpha }^3-\sin {\alpha }^2}{\alpha }$$$$=\frac{\sin {\alpha }^3-\sin {\alpha }^2}{\alpha }+\frac{2\sin {\alpha }^3-\sin {\alpha }^2}{\alpha }$$$$=\frac{3\sin {\alpha }^3-2\sin {\alpha }^2}{\alpha }$$

Answered by mathmax by abdo last updated on 02/May/21

$$\mathrm{\Phi }\left(\alpha \right)={\int }_{\alpha }^{{\alpha }^2}\frac{\sin \left(\alpha \mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\Rightarrow \mathrm{\Phi }\left(\alpha \right){=}_{\alpha \mathrm{x}=\mathrm{t}}{\int }_{{\alpha }^2}^{{\alpha }^3}\frac{\mathrm{sint}}{\frac{\mathrm{t}}{\alpha }}\frac{\mathrm{dt}}{\alpha }$$$$={\int }_{{\alpha }^2}^{{\alpha }^3}\frac{\mathrm{sint}}{\mathrm{t}}\mathrm{dt}\Rightarrow {\mathrm{\Phi }}^{{}^{\prime }}\left(\alpha \right)=\left(3{\alpha }^2\right)\frac{\sin \left({\alpha }^3\right)}{{\alpha }^3}-\left(2\alpha \right)\frac{\sin \left({\alpha }^2\right)}{{\alpha }^2}$$$$=\frac{3\sin \left({\alpha }^3\right)}{\alpha }-\frac{2\sin \left({\alpha }^2\right)}{\alpha }=\frac{3\sin \left({\alpha }^3\right)-2\sin \left({\alpha }^2\right)}{\alpha }$$

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