0-1-dx-ln-x-by-Gamma-function-

Question Number 97057 by bemath last updated on 06/Jun/20

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{dx}}{\:\sqrt{−\mathrm{ln}\left({x}\right)}}\:?\:\left[\:{by}\:{G}\mathrm{amma}\:\mathrm{function}\:\right] \\ $$

Answered by Sourav mridha last updated on 06/Jun/20

let ln(x)=−k.. =∫_0 ^∞ e^(−k) .k^(−(1/2)) dk =𝚪((1/2))=(√π). it is need not to tell −′′by Gamma function′′.

$$\boldsymbol{{let}}\:\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)=−\boldsymbol{{k}}.. \\ $$$$=\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{−\boldsymbol{{k}}} .\boldsymbol{{k}}^{−\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{{dk}} \\ $$$$=\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}. \\ $$$$\boldsymbol{{it}}\:\boldsymbol{{is}}\:\boldsymbol{{need}}\:\boldsymbol{{not}}\:\boldsymbol{{to}}\:\boldsymbol{{tell}}\:−''\boldsymbol{{by}}\:\boldsymbol{{Gamma}}\: \\ $$$$\boldsymbol{{function}}''. \\ $$

Commented by bemath last updated on 06/Jun/20

$$\mathrm{yes}.\:\mathrm{thanks} \\ $$

Answered by mathmax by abdo last updated on 06/Jun/20

I =∫_0 ^1 (dx/( (√(−lnx)))) changement (√(−lnx))=t give −lnx =t^2 ⇒lnx =−t^2 ⇒x =e^(−t^2 ) I =−∫_0 ^∞ ((−2t e^(−t^2 ) )/t) dt =2 ∫_0 ^∞ e^(−t^2 ) dt =2×((√π)/2) =(√π)

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\:\sqrt{−\mathrm{lnx}}}\:\mathrm{changement}\:\sqrt{−\mathrm{lnx}}=\mathrm{t}\:\mathrm{give}\:−\mathrm{lnx}\:=\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{lnx}\:=−\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{x}\:=\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \\ $$$$\mathrm{I}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{−\mathrm{2t}\:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } }{\mathrm{t}}\:\mathrm{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\:=\mathrm{2}×\frac{\sqrt{\pi}}{\mathrm{2}}\:=\sqrt{\pi} \\ $$

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