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Question Number 139851 by mnjuly1970 last updated on 01/May/21

$$$$$$prove......$$$$\mathrm{\Phi }={\int }_0^1\frac{ln\left(\frac{1}{1-x}\right)}{\sqrt{x}}dx=4ln\left(\frac{e}{2}\right)...✓$$

Answered by mindispower last updated on 01/May/21

$$\mathrm{\Phi }=-{\int }_0^1\frac{ln(1-x)}{\sqrt{x}}\Rightarrow \mathrm{\Phi }=-2{\int }_0^{1}ln(1-x).d\sqrt{x}$$$$=-2{\int }_0^{1}ln(1-t^2)dt=\underset{x\to 1}{\lim }{[-2tln(1-t^2)]}_0^x$$$$-4{\int }_0^x\frac{t^2}{1-t^2}=\underset{x\to 1}{\lim }-2xln(1-x)+4x-2{\int }_0^x\frac{1}{1-t}+\frac{1}{1+t}dt$$$$=\underset{x\to 1}{\lim }-2xln(1-x)+2ln(1-x)+4x-2ln(1+x)$$$$=\underset{x\to 1}{\lim 2}(1-x)ln(1-x)+4-2ln\left(2\right)$$$$4ln\left(e\right)-4ln\sqrt{2}=4ln\left(\frac{e}{\sqrt{2}}\right)$$$$$$

Answered by mnjuly1970 last updated on 01/May/21

Answered by mnjuly1970 last updated on 01/May/21

$$solution\left(2\right)$$$$\mathrm{\Phi }={\int }_0^1\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{n-\frac{1}{2}}}{n}dx=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n(n+\frac{1}{2})}$$$$=2\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n}-\frac{1}{n+\frac{1}{2}})=2(\gamma -\gamma +\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{1}{n+\frac{1}{2}})$$$$=2\gamma +2\psi \left(\frac{3}{2}\right)=2\gamma +2(2-\gamma -2ln\left(2\right))$$$$=4-4ln\left(2\right)=4ln\left(\frac{e}{2}\right).....$$$$$$

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