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Question Number 149100 by Naser last updated on 02/Aug/21

Answered by Kamel last updated on 02/Aug/21

$$1/\mathcal{L}\left(J_0\left(t\right)\right)=\frac{1}{\pi }{\int }_0^{\pi }{\int }_0^{+\mathrm{\infty }}cos\left(tsin\left(\theta \right)\right)e^{-st}dtd\theta =\frac{1}{\pi }{\int }_0^{\pi }\frac{sd\theta }{{sin}^2\left(\theta \right)+s^2}=\frac{2s}{\pi }{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{s^2+{cos}^2\left(\theta \right)}$$$$=\frac{2s}{\pi }{\int }_0^{+\mathrm{\infty }}\frac{du}{1+s^2+s^2{u}^2}=\frac{1}{\sqrt{1+s^2}}$$$$2/\beta (n,n)=\frac{\mathrm{\Gamma }\left(n\right)\mathrm{\Gamma }\left(n\right)}{\mathrm{\Gamma }\left(2n\right)}=\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(n\right)2^{1-2n}\frac{\mathrm{\Gamma }\left(n\right)2^{2n-1}}{\mathrm{\Gamma }\left(2n\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}=2^{1-2n}\frac{\mathrm{\Gamma }\left(n\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(n+\frac{1}{2})}=2^{1-2n}\beta (n,\frac{1}{2})$$$$3/{\mathcal{L}}^{-1}\left(\frac{d}{ds}{\mathcal{L}}^{-1}\left(Ln(1+\frac{1}{s^2})\right)\right)={\mathcal{L}}^{-1}(\frac{2s}{1+s^2}-\frac{2}{s})=2(cos\left(t\right)-1)$$$$\therefore {\mathcal{L}}^{-1}\left(Ln(1+\frac{1}{s^2})\right)=2\frac{1-cos\left(t\right)}{t}$$$$4/???$$$$5/{\int }_0^2{x}^3{J}_0\left(x\right)dx={\int }_0^2{x}^2\left({xJ}_{1-1}\left(x\right)\right)dx=8J_1\left(2\right)-2{\int }_0^2{x}^2{J}_{2-1}\left(x\right)dx=8(J_1\left(2\right)-J_2\left(2\right))$$$$6-1/{\int }_0^1{x}^m{Ln}^n\left(x\right)dx\stackrel{x=e^{-t}}{=}{(-1)}^n{\int }_0^{+\mathrm{\infty }}{t}^n{e}^{-(m+1)t}dt=\frac{{(-1)}^{n}n!}{{(m+1)}^{m+1}}$$$$6-2/{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{2\theta }d\theta }{e^{3\theta }+1}\stackrel{t=e^{-3\theta }}{=}\frac{1}{3}{\int }_0^{+\mathrm{\infty }}\frac{t^{-\frac{2}{3}}dt}{1+t}=\frac{2\pi }{3\sqrt{3}}$$$$7/\mathcal{L}\left(sin\left(\sqrt{t}\right)\right)={\int }_0^{+\mathrm{\infty }}sin\left(\sqrt{t}\right)e^{-st}dt\stackrel{u=\sqrt{t}}{=}2{\int }_0^{+\mathrm{\infty }}usin\left(u\right)e^{-{su}^2}du=\frac{1}{s}{\int }_0^{+\mathrm{\infty }}cos\left(u\right)e^{-{su}^2}du$$$$I\left(\alpha \right)={\int }_0^{+\mathrm{\infty }}cos\left(\alpha x\right)e^{-{sx}^2}dx=\frac{2s}{\alpha }{\int }_0^{+\mathrm{\infty }}xsin\left(\alpha x\right)e^{-{sx}^2}dx=-\frac{2s}{\alpha }{I}^{\prime }\left(\alpha \right)$$$$I\left(\alpha \right)={Ae}^{-\frac{1}{4s}{\alpha }^2},I\left(0\right)=\frac{\sqrt{\pi }}{2\sqrt{s}}\Rightarrow I\left(\alpha \right)=\frac{\sqrt{\pi }}{2\sqrt{s}}{e}^{-\frac{1}{4s}{\alpha }^2}$$$$\therefore \mathcal{L}\left(sin\left(\sqrt{t}\right)\right)=\frac{\sqrt{\pi }}{2s^{\frac{3}{2}}}{e}^{-\frac{1}{4s}}$$

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