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a-1-lnx-lny-1-dxdy-




Question Number 5058 by Yozzii last updated on 06/Apr/16
∫∫(a/((1−lnx)(lny−1)))dxdy=?
$$\int\int\frac{{a}}{\left(\mathrm{1}−{lnx}\right)\left({lny}−\mathrm{1}\right)}{dxdy}=? \\ $$
Commented by Yozzii last updated on 06/Apr/16
∫∫∫∫...∫∫∫(a/(Π_(i=1) ^n (1−lnx_i )))dx_1 dx_2 ...dx_n =?
$$\int\int\int\int…\int\int\int\frac{{a}}{\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\mathrm{1}−{lnx}_{{i}} \right)}{dx}_{\mathrm{1}} {dx}_{\mathrm{2}} …{dx}_{{n}} =? \\ $$
Commented by prakash jain last updated on 07/Apr/16
=∫(1/(1−ln x))dx∙∫(1/(1−ln y))dy  I don′t think ∫(1/(1−ln x))dx can be expressed  in terms of elementry functions.
$$=\int\frac{\mathrm{1}}{\mathrm{1}−\mathrm{ln}\:{x}}\mathrm{d}{x}\centerdot\int\frac{\mathrm{1}}{\mathrm{1}−\mathrm{ln}\:{y}}\mathrm{d}{y} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\int\frac{\mathrm{1}}{\mathrm{1}−\mathrm{ln}\:{x}}\mathrm{d}{x}\:\mathrm{can}\:\mathrm{be}\:\mathrm{expressed} \\ $$$$\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{elementry}\:\mathrm{functions}. \\ $$
Commented by Yozzii last updated on 07/Apr/16
Wolfram alpha defines the En function  for n=1 as E_1 (x)=∫_1 ^∞ (e^(−tx) /t)dt=∫_x ^∞ (e^(−u) /u)du. Now, define the  exponential integral Ei(x) such that  E_1 (x)=−Ei(−x). Then  Ei(x)=−∫_(−x) ^∞ (e^(−t) /t)dt.  Let u=1−lnx⇒du=−x^(−1) dx  x=e^(1−u) ⇒−e^(1−u) du=dx.  ∫(dx/(1−lnx))=∫((−e^(1−u) du)/u)=−e∫(e^(−u) /u)du  ∫(dx/(1−lnx))=eEi(u)+c=eEi(−(−u))+c=−eE_1 (−u)+c=−eE_1 (lnx−1)+c.  I′m not certain whether or not   I′m using this definition correctly.
$${Wolfram}\:{alpha}\:{defines}\:{the}\:{En}\:{function} \\ $$$${for}\:{n}=\mathrm{1}\:{as}\:{E}_{\mathrm{1}} \left({x}\right)=\overset{\infty} {\int}_{\mathrm{1}} \frac{{e}^{−{tx}} }{{t}}{dt}=\int_{{x}} ^{\infty} \frac{{e}^{−{u}} }{{u}}{du}.\:{Now},\:{define}\:{the} \\ $$$${exponential}\:{integral}\:{Ei}\left({x}\right)\:{such}\:{that} \\ $$$${E}_{\mathrm{1}} \left({x}\right)=−{Ei}\left(−{x}\right).\:{Then} \\ $$$${Ei}\left({x}\right)=−\int_{−{x}} ^{\infty} \frac{{e}^{−{t}} }{{t}}{dt}. \\ $$$${Let}\:{u}=\mathrm{1}−{lnx}\Rightarrow{du}=−{x}^{−\mathrm{1}} {dx} \\ $$$${x}={e}^{\mathrm{1}−{u}} \Rightarrow−{e}^{\mathrm{1}−{u}} {du}={dx}. \\ $$$$\int\frac{{dx}}{\mathrm{1}−{lnx}}=\int\frac{−{e}^{\mathrm{1}−{u}} {du}}{{u}}=−{e}\int\frac{{e}^{−{u}} }{{u}}{du} \\ $$$$\int\frac{{dx}}{\mathrm{1}−{lnx}}={eEi}\left({u}\right)+{c}={eEi}\left(−\left(−{u}\right)\right)+{c}=−{eE}_{\mathrm{1}} \left(−{u}\right)+{c}=−{eE}_{\mathrm{1}} \left({lnx}−\mathrm{1}\right)+{c}. \\ $$$${I}'{m}\:{not}\:{certain}\:{whether}\:{or}\:{not}\: \\ $$$${I}'{m}\:{using}\:{this}\:{definition}\:{correctly}. \\ $$$$ \\ $$

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