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Question Number 38463 by maxmathsup by imad last updated on 25/Jun/18

$$calculateI={\int }_0^1\frac{ln(1-\frac{t^2}{4})}{t^2}dt$$

Commented by abdo mathsup 649 cc last updated on 05/Jul/18

$$wehaveproved?that$$$${\int }_0^1\frac{ln(1-x^2{t}^2)}{t^2}dt=-ln(1-x^2)-\frac{x}{2}ln\left(\frac{1+x}{1-x}\right)$$$$with\mid x\mid <1\Rightarrow$$$${\int }_0^1\frac{ln(1-\frac{t^2}{4})}{t^2}dt=-ln(1-{\left(\frac{1}{2}\right)}^2)-\frac{1}{4}ln\left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)$$$$=-ln\left(\frac{3}{4}\right)-\frac{1}{4}ln\left(3\right)=ln\left(\frac{4}{3}\right)-\frac{ln\left(3\right)}{4}$$$$=2ln\left(2\right)-ln\left(3\right)-\frac{1}{4}ln\left(3\right)$$$$=2ln\left(2\right)-\frac{5}{4}ln\left(3\right).$$

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