Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 143454 by mnjuly1970 last updated on 14/Jun/21

$$$$$$........nice.......integral.......$$$$\mathcal{T}:={\int }_0^{\mathrm{\infty }}\frac{arctan\left(x\right)}{x^{ln\left(x\right)+1}}dx\stackrel{?}{=}\frac{\pi \sqrt{\pi }}{4}$$$$$$

Answered by mindispower last updated on 14/Jun/21

$$x\to \frac{1}{x}\Rightarrow$$$$A={\int }_0^{\mathrm{\infty }}\frac{\frac{\pi }{2}-arctan\left(x\right)}{{\left(\frac{1}{x}\right)}^{-ln\left(x\right)+1}}.\frac{dx}{x^2}=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}\frac{dx}{x^{ln\left(x\right)+1}}-{\int }_0^{\mathrm{\infty }}\frac{arctan\left(x\right)dx}{x^{ln\left(x\right)+1}}$$$$2A=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}\frac{dx}{x^{ln\left(x\right)+1}}$$$$ln\left(x\right)=u$$$$A=\frac{\pi }{4}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{du}{e^{u^2}}=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}{e}^{-u^2}du=\frac{\pi }{4}{\int }_0^{\mathrm{\infty }}{e}^{-t}.t^{-\frac{1}{2}}dt$$$$=\frac{\pi }{4}\mathrm{\Gamma }\left(\frac{1}{2}\right)=\frac{\pi }{4}\sqrt{\pi }$$$${\int }_0^{\mathrm{\infty }}\frac{arctan\left(x\right)}{x^{ln\left(x\right)+1}}dx=\frac{\pi \sqrt{\pi }}{4}$$$$$$

Commented by mnjuly1970 last updated on 14/Jun/21

$$thankdalot..$$

Commented by mindispower last updated on 14/Jun/21

$$pleasur$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com