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

$$\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{−{x}} \left({x}^{\mathrm{10}} −\mathrm{1}\right)}{{ln}\left({x}\right)}\:{dx}\: \\$$

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

$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} \left({x}^{{a}} −\mathrm{1}\right)}{{lnx}}{dx} \\$$$$\frac{{dI}\left({a}\right)}{{da}}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} ×{x}^{{a}} {lnx}}{{lnx}}{dx} \\$$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{a}+\mathrm{1}−\mathrm{1}} {dx} \\$$$$=\lceil\left({a}+\mathrm{1}\right)\leftarrow{gamma}\:{function} \\$$$$={a}! \\$$$${dI}\left({a}\right)={a}!{da} \\$$$${I}\left({a}\right)=\int\lceil\left({a}+\mathrm{1}\right){da} \\$$

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

$${Sir},\:{there}\:{is}\:{a}\:{option}\:{for}\:{Gammafunction} \\$$$$\left.{Press}\:'\boldsymbol{\alpha}'\:{you}\:{will}\:{be}\:{able}\:{to}\:{see}\::\right) \\$$$$'\Gamma''\Lambda''... \\$$

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

$${ok}\:{thank}\:{you} \\$$

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

$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} \left({x}^{{a}} −\mathrm{1}\right)}{{log}\left({x}\right)}{dx} \\$$$${I}'\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{a}} {dx} \\$$$${I}'\left({a}\right)=\Gamma\left({a}+\mathrm{1}\right) \\$$$${I}\left({a}\right)=\int\Gamma\left({a}+\mathrm{1}\right){da}... \\$$