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0-e-x-x-10-1-ln-x-dx-




Question Number 119291 by 675480065 last updated on 23/Oct/20
∫_0 ^( ∞) ((e^(−x) (x^(10) −1))/(ln(x))) dx
$$\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{−{x}} \left({x}^{\mathrm{10}} −\mathrm{1}\right)}{{ln}\left({x}\right)}\:{dx}\: \\ $$
Answered by TANMAY PANACEA last updated on 23/Oct/20
I(a)=∫_0 ^∞ ((e^(−x) (x^a −1))/(lnx))dx  ((dI(a))/da)=∫_0 ^∞ ((e^(−x) ×x^a lnx)/(lnx))dx  =∫_0 ^∞ e^(−x) x^(a+1−1) dx  =⌈(a+1)←gamma function  =a!  dI(a)=a!da  I(a)=∫⌈(a+1)da
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} \left({x}^{{a}} −\mathrm{1}\right)}{{lnx}}{dx} \\ $$$$\frac{{dI}\left({a}\right)}{{da}}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} ×{x}^{{a}} {lnx}}{{lnx}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{a}+\mathrm{1}−\mathrm{1}} {dx} \\ $$$$=\lceil\left({a}+\mathrm{1}\right)\leftarrow{gamma}\:{function} \\ $$$$={a}! \\ $$$${dI}\left({a}\right)={a}!{da} \\ $$$${I}\left({a}\right)=\int\lceil\left({a}+\mathrm{1}\right){da} \\ $$
Commented by Dwaipayan Shikari last updated on 23/Oct/20
Sir, there is a option for Gammafunction  Press ′𝛂′ you will be able to see :)  ′Γ′′Λ′′...
$${Sir},\:{there}\:{is}\:{a}\:{option}\:{for}\:{Gammafunction} \\ $$$$\left.{Press}\:'\boldsymbol{\alpha}'\:{you}\:{will}\:{be}\:{able}\:{to}\:{see}\::\right) \\ $$$$'\Gamma''\Lambda''… \\ $$
Commented by TANMAY PANACEA last updated on 23/Oct/20
ok thank you
$${ok}\:{thank}\:{you} \\ $$
Answered by Dwaipayan Shikari last updated on 23/Oct/20
I(a)=∫_0 ^∞ ((e^(−x) (x^a −1))/(log(x)))dx  I′(a)=∫_0 ^∞ e^(−x) x^a dx  I′(a)=Γ(a+1)  I(a)=∫Γ(a+1)da...
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} \left({x}^{{a}} −\mathrm{1}\right)}{{log}\left({x}\right)}{dx} \\ $$$${I}'\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{a}} {dx} \\ $$$${I}'\left({a}\right)=\Gamma\left({a}+\mathrm{1}\right) \\ $$$${I}\left({a}\right)=\int\Gamma\left({a}+\mathrm{1}\right){da}… \\ $$

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