Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 119291 by 675480065 last updated on 23/Oct/20

$${\int }_0^{\mathrm{\infty }}\frac{e^{-x}(x^{10}-1)}{ln\left(x\right)}dx$$

Answered by TANMAY PANACEA last updated on 23/Oct/20

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-x}(x^a-1)}{lnx}dx$$$$\frac{dI\left(a\right)}{da}={\int }_0^{\mathrm{\infty }}\frac{e^{-x}\times x^{a}lnx}{lnx}dx$$$$={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{a+1-1}dx$$$$=\lceil (a+1)\leftarrow gammafunction$$$$=a!$$$$dI\left(a\right)=a!da$$$$I\left(a\right)=\int \lceil (a+1)da$$

Commented by Dwaipayan Shikari last updated on 23/Oct/20

$$Sir,thereisaoptionforGammafunction$$$$Press{}^{\prime }{\mathit{\alpha }}^{\prime }youwillbeabletosee:)$$$${}^{\prime }{\mathrm{\Gamma }}^{″}{\mathrm{\Lambda }}^{″}...$$

Commented by TANMAY PANACEA last updated on 23/Oct/20

$$okthankyou$$

Answered by Dwaipayan Shikari last updated on 23/Oct/20

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-x}(x^a-1)}{log\left(x\right)}dx$$$$I^{\prime }\left(a\right)={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{a}dx$$$$I^{\prime }\left(a\right)=\mathrm{\Gamma }(a+1)$$$$I\left(a\right)=\int \mathrm{\Gamma }(a+1)da...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com