Question Number 96990 by Ar Brandon last updated on 06/Jun/20
$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sqrt{\mathrm{x}}\right)^{\sqrt{\mathrm{x}}} \mathrm{dx} \\ $$
Answered by Sourav mridha last updated on 06/Jun/20
![∫_0 ^1 e^(((√x)/2).ln(x)) dx=Σ_(n=0) ^∞ (1/2^n )∫_0 ^∞ ((x^(n/2) .[ln(x)]^n )/(n!))dx now let ln(x)=−k so we get =Σ_(n=0) ^∞ (((−1)^n )/(2^n n!))∫_0 ^∞ e^(−((n/2)+1)k) .k^n dk =Σ_(n=0) ^∞ (((−1)^n )/(2^n 𝚪(n+1))).((𝚪(n+1))/(((n/2)+1)^(n+1) )).[using Laplace transform =2Σ_(n=0) ^∞ (((−1)^n )/((n+2)^(n+1) )).](https://www.tinkutara.com/question/Q97014.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{e}}^{\frac{\sqrt{\boldsymbol{{x}}}}{\mathrm{2}}.\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)} \boldsymbol{{dx}}=\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\boldsymbol{{n}}} }\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{x}}^{\frac{\boldsymbol{{n}}}{\mathrm{2}}} .\left[\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)\right]^{\boldsymbol{{n}}} }{\mathrm{n}!}\boldsymbol{{dx}} \\ $$$$\:\:\:\:\boldsymbol{{now}}\:\boldsymbol{{let}}\:\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)=−\boldsymbol{{k}}\:\boldsymbol{{so}}\:\boldsymbol{{we}}\:\boldsymbol{{get}} \\ $$$$=\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\mathrm{2}^{\boldsymbol{{n}}} \boldsymbol{{n}}!}\int_{\mathrm{0}} ^{\infty} \boldsymbol{{e}}^{−\left(\frac{\boldsymbol{{n}}}{\mathrm{2}}+\mathrm{1}\right)\boldsymbol{{k}}} .\boldsymbol{{k}}^{\boldsymbol{{n}}} \boldsymbol{{dk}} \\ $$$$=\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2}^{\boldsymbol{{n}}} \boldsymbol{\Gamma}\left(\boldsymbol{{n}}+\mathrm{1}\right)}.\frac{\boldsymbol{\Gamma}\left(\boldsymbol{{n}}+\mathrm{1}\right)}{\left(\frac{\boldsymbol{{n}}}{\mathrm{2}}+\mathrm{1}\right)^{\boldsymbol{{n}}+\mathrm{1}} }.\left[\boldsymbol{{using}}\:\boldsymbol{{L}}\mathrm{aplace}\:\boldsymbol{{transform}}\right. \\ $$$$=\mathrm{2}\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\left(\boldsymbol{{n}}+\mathrm{2}\right)^{\boldsymbol{{n}}+\mathrm{1}} }. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Ar Brandon last updated on 12/Jun/20
��thanks
Commented by smridha last updated on 12/Jun/20
$$\boldsymbol{{welcome}} \\ $$