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Question Number 115222 by mnjuly1970 last updated on 24/Sep/20

$$....nicemath...$$$$$$$$niceintegral$$$$prove::$$$$\mathrm{\Psi }=9{\int }_0^{\mathrm{\infty }}{x}^5{e}^{-x^3}ln(1+x)dx\stackrel{???}{=}\mathrm{\Gamma }\left(\frac{1}{3}\right)-\mathrm{\Gamma }\left(\frac{2}{3}\right)+\mathrm{\Gamma }\left(\frac{3}{3}\right)$$$$m.n.july.1970$$$$$$

Answered by mathdave last updated on 24/Sep/20

$$solution$$$$letI=9{\int }_0^{\mathrm{\infty }}{x}^5{e}^{-x^3}\ln (1+x)dx$$$$let\int dv=\int x^5{e}^{-x^3}dxputy=x^3$$$$V=\frac{1}{3}\int {ye}^{-y}dybyIBP$$$$V=-\frac{1}{3}{ye}^{-y}+\int e^{-y}dy=-\frac{1}{3}{e}^{-y}(y+1)+kbuty=x^3$$$$\because V=-\frac{1}{3}{e}^{-x^3}(x^3+1)+kandu=\ln (1+x),du=\frac{1}{1+x}$$$$usingIBP\int udv=uv-\int vdu$$$$I=9{(-\frac{1}{3}{e}^{-x^3}(x^3+1)\ln (1+x))}_0^{\mathrm{\infty }}+\frac{9}{3}{\int }_0^{\mathrm{\infty }}\left(\frac{1+x^3}{1+x}\right)e^{-x^3}dx$$$$I=0+3{\underset{0}{\int }}^{\mathrm{\infty }}(x^2-x+1)e^{-x^3}dxlet$$$$y=x^3,x^{\frac{1}{3}}anddx=\frac{1}{3}{y}^{-\frac{2}{3}}dy$$$$I=3{\int }_0^{\mathrm{\infty }}(y^{\frac{2}{3}}-y^{\frac{1}{3}}+1)e^{-y}\times \frac{1}{3}{y}^{-\frac{2}{3}}dy$$$$I={\int }_0^{\mathrm{\infty }}(1-y^{-\frac{1}{3}}+y^{-\frac{2}{3}})e^{-y}dy$$$$I={\int }_0^{\mathrm{\infty }}{y}^{1-1}{e}^{-y}dy-{\int }_0^{\mathrm{\infty }}{y}^{\frac{2}{3}-1}{e}^{-y}dy+{\int }_0^{\mathrm{\infty }}{y}^{\frac{1}{3}-1}{e}^{-y}dy$$$$I=\mathrm{\Gamma }\left(1\right)-\mathrm{\Gamma }\left(\frac{2}{3}\right)+\mathrm{\Gamma }\left(\frac{1}{3}\right)$$$$\because {\int }_0^{\mathrm{\infty }}{x}^5{e}^{-x^3}\ln (1+x)dx=\mathrm{\Gamma }\left(\frac{1}{3}\right)-\mathrm{\Gamma }\left(\frac{2}{3}\right)+\mathrm{\Gamma }\left(\frac{3}{3}\right)Q.E.D$$$$bymathdave\left(24/09/2020\right)$$$$$$

Commented by mnjuly1970 last updated on 24/Sep/20

$$thankyou$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

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