Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 123261 by mnjuly1970 last updated on 24/Nov/20

$$...nicecalculus...$$$$provethat::$$$$$$$$\mathrm{\Omega }={\int }_{\mathbb{R}}{e}^{x-{sinh}^2\left(x\right)}dx=\sqrt{\pi }$$

Answered by Olaf last updated on 24/Nov/20

$$\mathrm{\Omega }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^x{e}^{-{\sinh }^{2}x}dx\left(1\right)$$$$\mathrm{Let}u=-x$$$$\mathrm{\Omega }={\int }_{+\mathrm{\infty }}^{-\mathrm{\infty }}{e}^{-x}{e}^{-{\sinh }^{2}x}(-dx)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-x}{e}^{-{\sinh }^{2}x}dx\left(2\right)$$$$\left(1\right)+\left(2\right):$$$$2\mathrm{\Omega }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}(e^x+e^{-x})e^{-{\sinh }^{2}x}dx$$$$\Rightarrow \mathrm{\Omega }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\cosh {xe}^{-{\sinh }^{2}x}dx$$$$\Rightarrow \mathrm{\Omega }=2{\int }_0^{+\mathrm{\infty }}\cosh {xe}^{-{\sinh }^{2}x}dx$$$$\mathrm{Let}u=\sinh x,du=\cosh xdx$$$$\mathrm{\Omega }=2{\int }_0^{+\mathrm{\infty }}{e}^{-u^2}du=2\times {\mathrm{erf}}_{\mathrm{\infty }}=2\times \frac{\sqrt{\pi }}{2}$$$$\mathrm{\Omega }=\sqrt{\pi }$$

Commented by mnjuly1970 last updated on 24/Nov/20

$$bravo$$$$excellent.thankyoumaster$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com