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Question Number 137474 by mnjuly1970 last updated on 03/Apr/21

$$.....nice........calculus.....$$$$\mathit{\varphi }={\int }_0^{1}ln\left(\sqrt[3]{ln\left(\sqrt{1-x}\right)}\right)$$$$Im\left(\mathit{\varphi }\right)=???$$

Answered by mindispower last updated on 03/Apr/21

$$Im\left(\varphi \right)=\frac{1}{3}{\int }_0^{1}ln\left(ln\left(x\right)\right)dx$$$$=Im\frac{1}{3}{\int }_{-\mathrm{\infty }}^{0}ln\left(t\right)e^{t}dt$$$$=Im\frac{1}{3}{\int }_0^{\mathrm{\infty }}ln(-x)e^{-x}dx$$$$=\frac{1\pi }{3}{\int }_0^{\mathrm{\infty }}{e}^{-t}dt=\frac{\pi }{3}\mathrm{\Gamma }\left(1\right)=\frac{\pi }{3}$$

Commented by mnjuly1970 last updated on 03/Apr/21

$$verynice..thankyoumrpower...$$

Answered by Ñï= last updated on 03/Apr/21

$${\int }_0^{1}ln\left(\sqrt[3]{ln\left(\sqrt{1-x}\right)}\right)=\frac{1}{3}{\int }_0^{1}ln\left(ln\sqrt{x}\right)dx=\frac{1}{3}{\int }_0^{1}ln\left(\frac{1}{2}lnx\right)dx$$$$=\frac{1}{3}{\int }_0^1(-ln2+lnlnx)dx=-\frac{1}{3}ln2+\frac{1}{3}{\int }_0^{1}lnlnxdx$$$$=-\frac{1}{3}ln2+\frac{1}{3}{\int }_{-\mathrm{\infty }}^0{e}^{u}lnudu$$$$=-\frac{1}{3}ln2+\frac{1}{3}{\int }_0^{\mathrm{\infty }}{e}^{-u}ln(-u)du$$$$=-\frac{1}{3}ln2+\frac{1}{3}{\int }_0^{\mathrm{\infty }}{e}^{-u}ln\left(e^{i\pi }u\right)du$$$$=-\frac{1}{3}ln2+\frac{1}{3}{\int }_0^{\mathrm{\infty }}{e}^{-u}(lnu+i\pi )du$$$$=-\frac{1}{3}ln2+\frac{1}{3}\frac{\partial }{\partial a}{\mid }_{a=0}{\int }_0^{\mathrm{\infty }}{u}^a{e}^{-u}du+\frac{i\pi }{3}$$$$=-\frac{1}{3}ln2+\frac{1}{3}\frac{\partial }{\partial a}{\mid }_{a=0}\mathrm{\Gamma }(a+1)+\frac{i\pi }{3}$$$$=-\frac{1}{3}ln2+\frac{1}{3}\psi \left(1\right)+\frac{i\pi }{3}$$$$im\left(\varphi \right)=\frac{\pi }{3}$$$$------------$$$${\int }_0^1{(-1)}^{x}dx={\int }_0^1{e}^{i\pi x}dx=\frac{2i}{\pi }$$

Commented by mnjuly1970 last updated on 03/Apr/21

$$thanksalot..$$

Answered by Dwaipayan Shikari last updated on 03/Apr/21

$$\frac{1}{3}{\int }_0^{1}log\left(\frac{1}{2}log\left(x\right)\right)dx$$$$=-\frac{1}{3}log\left(2\right)+\frac{1}{3}{\int }_0^{1}log\left(log\left(x\right)\right)dxx=e^{-t}\Rightarrow 1=-e^{-t}\frac{dt}{dx}$$$$=-\frac{1}{3}log\left(2\right)+\frac{1}{3}{\int }_0^{\mathrm{\infty }}{e}^{-t}log(-t)dt$$$$=-\frac{1}{3}log\left(2\right)+\frac{\pi i}{3}-\frac{\gamma }{3}$$$$Im\left(\varphi \right)=\frac{\pi }{3}$$

Commented by Dwaipayan Shikari last updated on 03/Apr/21

$$Ifitwas{\int }_0^{1}log\left(\sqrt[3]{log\left(\sqrt{\frac{1}{1-x}}\right)}\right)dxthentheanswerwillbe$$$$=\frac{-1}{3}log\left(2\right)-\frac{\gamma }{3}$$

Commented by mnjuly1970 last updated on 03/Apr/21

$$gratefulmrpayan...$$

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