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Question Number 190692 by mnjuly1970 last updated on 09/Apr/23

$$$$$$\mathrm{calculate}$$$$$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}\frac{{\sin }^3\left(x\right)\ln \left(x\right)}{x}\mathrm{d}x=?$$$$$$$$@\mathrm{nice}-\mathrm{mathematics}$$$$$$

Answered by 07049753053 last updated on 09/Apr/23

$$weknowthat{sin}^3\left(x\right)=\frac{1}{4}(3sin\left(x\right)-sin\left(3x\right))$$$$\frac{3}{4}{\int }_0^{\mathrm{\infty }}\frac{\mathbf{sin}\left(\mathbf{x}\right)\mathbf{ln}\left(\mathbf{x}\right)}{\mathbf{x}}\mathbf{dx}-\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{\mathbf{sin}\left(3\mathbf{x}\right)\mathbf{ln}\left(\mathbf{x}\right)}{\mathbf{x}}\mathbf{dx}$$$$\frac{3}{4}\left(\frac{-\mathit{\gamma }\mathit{\pi }}{2}\right)-\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{\mathbf{sin}\left(3\mathbf{x}\right)\mathbf{ln}\left(\mathbf{x}\right)}{\mathbf{x}}\mathbf{dx}$$$$-\frac{3}{8}\mathit{\gamma }\mathit{\pi }-\frac{1}{4}\frac{\mathit{d}}{\mathit{d}\mathit{a}}{\mid }_{\mathit{a}=1}{\int }_0^{\mathrm{\infty }}\frac{\mathit{s}\mathit{i}\mathit{n}\left(\mathit{z}\mathit{x}\right){\mathit{x}}^{\mathit{a}}}{\mathit{x}}\mathit{d}\mathit{x}$$$$\frac{\mathit{d}}{\mathit{d}\mathit{a}}{\mid }_{\mathit{a}=1}{\int }_0^{\mathrm{\infty }}\mathbf{sin}\left(\mathbf{zx}\right){\mathbf{x}}^{\mathbf{a}-1}\mathbf{dx}$$$$\mathbf{let}\mathbf{zx}=\mathbf{u}\mathbf{dx}=\frac{\mathbf{du}}{\mathbf{z}}$$$$\frac{\mathbf{d}}{\mathbf{da}}{\mid }_{\mathbf{a}=1}\frac{1}{\mathbf{z}}{\int }_0^{\mathrm{\infty }}\mathbf{sin}\left(\mathbf{u}\right){\left(\frac{\mathbf{u}}{\mathbf{z}}\right)}^{\mathbf{a}-1}\mathbf{du}=\frac{\mathbf{d}}{\mathbf{da}}{\mid }_{\mathit{a}=1}\left[\frac{1}{{\mathit{z}}^{\mathit{a}}}{\int }_0^{\mathrm{\infty }}\mathit{s}\mathit{i}\mathit{n}\left(\mathit{u}\right){\mathit{u}}^{\mathit{a}-1}\mathit{d}\mathit{u}\right]$$$$\mathit{f}\mathit{r}\mathit{o}\mathit{m}\mathit{e}\mathit{u}\mathit{l}\mathit{e}\mathit{r}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{a}$$$$\mathit{s}\mathit{i}\mathit{n}\left(\mathit{x}\right)=\mathit{I}\mathit{m}\left({\mathit{e}}^{-\mathit{i}\mathit{x}}\right)$$$$\frac{\mathit{d}}{\mathit{d}\mathit{a}}{\mid }_{\mathit{a}=1}\left[\frac{1}{{\mathit{z}}^{\mathit{a}}}\mathit{I}\mathit{m}{\int }_0^{\mathrm{\infty }}{\mathit{e}}^{-\mathit{i}\mathit{u}}{\mathit{u}}^{\mathit{a}-1}\mathit{d}\mathit{u}\right]$$$$\mathit{l}\mathit{e}\mathit{t}\mathit{u}\mathit{i}=\mathit{k}\mathit{d}\mathit{u}=\frac{\mathit{d}\mathit{k}}{\mathit{i}}$$$$\frac{\mathit{d}}{\mathit{d}\mathit{a}}{\mid }_{\mathit{a}=1}\left[\frac{1}{{\mathit{z}}^{\mathit{a}}}\mathit{I}\mathit{m}{\int }_0^{\mathrm{\infty }}{\mathit{e}}^{-\mathit{k}}{\left(\frac{\mathit{k}}{\mathit{i}}\right)}^{\mathit{a}-1}\frac{\mathit{d}\mathit{k}}{\mathit{i}}\right]$$$$\frac{\mathit{d}}{\mathit{d}\mathit{a}}{\mid }_{\mathit{a}=1}\left[\frac{1}{{\mathit{z}}^{\mathit{a}}}\mathit{I}\mathit{m}\left(\frac{1}{\mathit{i}}\right)\mathbf{\Gamma }\left(\mathit{a}\right)\right]=\frac{\mathit{d}}{\mathit{d}\mathit{a}}{\mid }_{\mathit{a}=1}\left[\frac{-\mathbf{\Gamma }\left(\mathit{a}\right)\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\pi }\mathit{a}}{2}\right)}{{\mathit{z}}^{\mathit{a}}}\right]$$$${[-\frac{z^{-a}}{2}\mathbf{\Gamma }\left(\mathit{a}\right)(2\mathit{\pi }\mathit{s}\mathit{i}\mathit{n}\left(\frac{\mathit{\pi }\mathit{a}}{2}\right)(\mathit{l}\mathit{o}\mathit{g}\left(\mathit{z}\right)-\mathit{\psi }\left(\mathit{a}\right))-\mathit{\pi }\mathit{c}\mathit{o}\mathit{s}\left(\frac{\mathit{\pi }\mathit{a}}{2}\right))]}_{\mathit{a}=1}$$$$-\frac{{\mathit{z}}^{-1}}{2}[2\mathit{\pi }(\mathit{l}\mathit{o}\mathit{g}\left(\mathit{z}\right)+\mathit{\gamma })=\frac{\mathit{\pi }}{\mathit{z}}\mathit{l}\mathit{o}\mathit{g}\left(\mathit{z}\right)+\frac{\mathit{\gamma }}{\mathit{z}}\mathit{\pi }$$$$\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathit{z}=3$$$$(-\frac{1}{4})(\frac{\mathit{\pi }}{3}\mathit{l}\mathit{o}\mathit{g}\left(3\right)+\frac{\mathit{\gamma }}{3}\mathit{\pi })-\frac{3}{8}\pi \mathit{\gamma }=-\frac{\mathit{\pi }}{12}\mathit{l}\mathit{o}\mathit{g}\left(3\right)-\frac{\mathit{\gamma }\mathit{\pi }}{12}-\frac{3\mathit{\pi }}{8}\gamma =\frac{\mathit{\pi }}{4}(\frac{\mathit{l}\mathit{o}\mathit{g}\left(3\right)}{3}-\frac{\mathit{\gamma }}{3}-\frac{3\mathit{\gamma }}{4})=\frac{\mathit{\pi }}{4}(\frac{\mathit{l}\mathit{o}\mathit{g}\left(3\right)}{3}-\frac{7\mathit{\gamma }}{12})=\frac{\mathit{\pi }}{12}(\mathit{l}\mathit{o}\mathit{g}\left(3\right)-\frac{7\mathit{\gamma }}{4})$$$$$$

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