Question Number 77790 by aliesam last updated on 10/Jan/20
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\sqrt{−{log}\left({x}\right)}}\:{dx} \\ $$
Answered by MJS last updated on 10/Jan/20
![∫(dx/(√(−ln x)))= [t=(√(−ln x)) → dx=−2x(√(−ln x))dt] =−2∫e^(−t^2 ) dt=−(√π)∫((2e^(−t^2 ) )/(√π))dt=−(√π)erf t = =−(√π)erf (√(−ln x)) +C ⇒ ∫_0 ^1 (dx/(√(−ln x)))=(√π)](Q77795.png)
$$\int\frac{{dx}}{\sqrt{−\mathrm{ln}\:{x}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{−\mathrm{ln}\:{x}}\:\rightarrow\:{dx}=−\mathrm{2}{x}\sqrt{−\mathrm{ln}\:{x}}{dt}\right] \\ $$$$=−\mathrm{2}\int\mathrm{e}^{−{t}^{\mathrm{2}} } {dt}=−\sqrt{\pi}\int\frac{\mathrm{2e}^{−{t}^{\mathrm{2}} } }{\sqrt{\pi}}{dt}=−\sqrt{\pi}\mathrm{erf}\:{t}\:= \\ $$$$=−\sqrt{\pi}\mathrm{erf}\:\sqrt{−\mathrm{ln}\:{x}}\:+{C} \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{\sqrt{−\mathrm{ln}\:{x}}}=\sqrt{\pi} \\ $$