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Question Number 116744 by mnjuly1970 last updated on 06/Oct/20

$$...\frac{nice}{calculus}...$$$$provethat::$$$$$$$${\int }_{-1}^{\mathrm{\infty }}\frac{e^{-4x}}{\sqrt{x+1}}dx=\frac{\sqrt{\pi }}{2}{e}^4$$$$...m.n.1970...$$$$$$

Answered by mindispower last updated on 06/Oct/20

$$x+1=t\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-4t+4}}{\sqrt{t}}dt$$$$4t=w\Rightarrow e^4{\int }_0^{\mathrm{\infty }}\frac{e^{-w}}{\sqrt{w}}.\frac{dw}{2}=\frac{e^4}{2}.{\int }_0^{\mathrm{\infty }}{w}^{\frac{1}{2}-1}{e}^{-w}dw$$$$=\frac{e^4}{2}\mathrm{\Gamma }\left(\frac{1}{2}\right)=\frac{e^4\sqrt{\pi }}{2}$$

Commented by mnjuly1970 last updated on 06/Oct/20

$$thankyousomuch...$$

Answered by Dwaipayan Shikari last updated on 06/Oct/20

$${\int }_{-1}^{\mathrm{\infty }}\frac{e^{-4x}}{\sqrt{1+x}}dx1+x=t^2\Rightarrow 1=2t\frac{dt}{dx}$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-4(t^2-1)}2t}{t}dt$$$$2e^4{\int }_0^{\mathrm{\infty }}{e}^{-4t^2}dt=2e^4\frac{1}{4}{\int }_0^{\mathrm{\infty }}{e}^{-u^2}du=\frac{\sqrt{\pi }}{2}{e}^4$$

Commented by mnjuly1970 last updated on 06/Oct/20

$$grateful..$$

Answered by 1549442205PVT last updated on 06/Oct/20

$$\mathrm{I}={\int }_{-1}^{\mathrm{\infty }}\frac{e^{-4x}}{\sqrt{x+1}}dx\stackrel{\mathrm{I}.\mathrm{B}.\mathrm{P}}{=}{2\mathrm{e}}^{-4\mathrm{x}}\sqrt{\mathrm{x}+1}{\mid }_{-1}^{+\mathrm{\infty }}$$$$+8{\int }_{-1}^{\mathrm{\infty }}{\mathrm{e}}^{-4\mathrm{x}}\sqrt{\mathrm{x}+1}\mathrm{dx}=8{\int }_{-1}^{\mathrm{\infty }}{\mathrm{e}}^{-4\mathrm{x}}\sqrt{\mathrm{x}+1}\mathrm{dx}$$$$\mathrm{Put}\sqrt{\mathrm{x}+1}=\mathrm{u}\Rightarrow \mathrm{x}+1={\mathrm{u}}^2\Rightarrow \mathrm{dx}=2\mathrm{udu}$$$$\mathrm{J}={\int }_{-1}^{\mathrm{\infty }}{\mathrm{e}}^{-4\mathrm{x}}\sqrt{\mathrm{x}+1}\mathrm{dx}=2{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-4({\mathrm{u}}^2-1)}{\mathrm{u}}^2\mathrm{du}$$$$\text{Missing left or extra right}$$$$\mathrm{Put}{\mathrm{v}}^2=\mathrm{t}\Rightarrow 2\mathrm{vdv}=\mathrm{dt}\Rightarrow \mathrm{dv}=\frac{\mathrm{dt}}{2\sqrt{\mathrm{t}}}$$$$\mathrm{J}=\frac{{\mathrm{e}}^4}{8}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}.{\mathrm{t}}^{\frac{1}{2}}\mathrm{dt}=\frac{{\mathrm{e}}^4}{8}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}.{\mathrm{t}}^{\frac{3}{2}-1}\mathrm{dt}$$$$=\frac{{\mathrm{e}}^4}{8}\mathrm{\Gamma }\left(\frac{3}{2}\right)=\frac{{\mathrm{e}}^4}{8}.\frac{\sqrt{\pi }}{2}.\mathrm{Therefore},$$$$\mathrm{I}=8\mathrm{J}=\frac{{\mathrm{e}}^4\sqrt{\pi }}{2}(\mathbf{Q}.\mathbf{E}.\mathbf{D})$$

Commented by mnjuly1970 last updated on 06/Oct/20

$$thankyou...$$

Commented by 1549442205PVT last updated on 07/Oct/20

$$\mathrm{Thank}\mathrm{Sir}.\mathrm{You}\mathrm{are}\mathrm{welcome}$$

Answered by mathmax by abdo last updated on 06/Oct/20

$${\int }_{-1}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-4\mathrm{x}}}{\sqrt{\mathrm{x}+1}}\mathrm{dx}{=}_{\sqrt{\mathrm{x}+1}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-4({\mathrm{t}}^2-1)}}{\mathrm{t}}\left(2\mathrm{t}\right)\mathrm{dt}=2{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{4\mathrm{t}}^2+4}\mathrm{dt}$$$$=2{\mathrm{e}}^4{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\left(2\mathrm{t}\right)}^2}\mathrm{dt}{=}_{2\mathrm{t}=\mathrm{z}}{2\mathrm{e}}^4{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{z}}^2}\frac{\mathrm{dz}}{2}={\mathrm{e}}^4{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}$$$$={\mathrm{e}}^4.\frac{\sqrt{\pi }}{2}=\frac{\sqrt{\pi }}{2}{\mathrm{e}}^4$$$$$$

Commented by mnjuly1970 last updated on 06/Oct/20

$$gratefulsir..$$$$veryniceasalways..$$

Commented by Bird last updated on 07/Oct/20

$$youarewelcomesir.$$

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