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Question Number 122412 by mohammad17 last updated on 16/Nov/20
∫∫((dxdy)/(1−x^2 y^2 ))
$$\int\int\frac{{dxdy}}{\mathrm{1}−{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$
Answered by Olaf last updated on 16/Nov/20
I(x,y) = ∫∫((dxdy)/(1−x^2 y^2 )) I(x,y) = ∫∫((dxdy)/((1−xy)(1+xy))) I(x,y) = (1/2)∫∫((1/(1−xy))+(1/(1+xy)))dxdy I(x,y) = (1/2)∫(−(1/y)ln∣1−xy∣+(1/y)ln∣1+xy∣)dy I(x,y) = −(1/2)∫(1/y)ln(((1−xy)/(1+xy)))dy Let u = xy I(x,y) = −(1/2)∫(x/u)ln(((1−u)/(1+u)))(du/x) I(x,y) = −(1/2)∫ln(((1−u)/(1+u)))(du/u) I(x,y) = (1/2)(dilog(1−u)−dilog(1+u)) I(x,y) = (1/2)(dilog(1−xy)−dilog(1+xy))
$$\mathrm{I}\left({x},{y}\right)\:=\:\int\int\frac{{dxdy}}{\mathrm{1}−{x}^{\mathrm{2}} {y}^{\mathrm{2}} } \\ $$$$\mathrm{I}\left({x},{y}\right)\:=\:\int\int\frac{{dxdy}}{\left(\mathrm{1}−{xy}\right)\left(\mathrm{1}+{xy}\right)} \\ $$$$\mathrm{I}\left({x},{y}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\int\left(\frac{\mathrm{1}}{\mathrm{1}−{xy}}+\frac{\mathrm{1}}{\mathrm{1}+{xy}}\right){dxdy} \\ $$$$\mathrm{I}\left({x},{y}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\left(−\frac{\mathrm{1}}{{y}}\mathrm{ln}\mid\mathrm{1}−{xy}\mid+\frac{\mathrm{1}}{{y}}\mathrm{ln}\mid\mathrm{1}+{xy}\mid\right){dy} \\ $$$$\mathrm{I}\left({x},{y}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{y}}\mathrm{ln}\left(\frac{\mathrm{1}−{xy}}{\mathrm{1}+{xy}}\right){dy} \\ $$$$\mathrm{Let}\:{u}\:=\:{xy} \\ $$$$\mathrm{I}\left({x},{y}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}}{{u}}\mathrm{ln}\left(\frac{\mathrm{1}−{u}}{\mathrm{1}+{u}}\right)\frac{{du}}{{x}} \\ $$$$\mathrm{I}\left({x},{y}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{ln}\left(\frac{\mathrm{1}−{u}}{\mathrm{1}+{u}}\right)\frac{{du}}{{u}} \\ $$$$\mathrm{I}\left({x},{y}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{dilog}\left(\mathrm{1}−{u}\right)−\mathrm{dilog}\left(\mathrm{1}+{u}\right)\right) \\ $$$$\mathrm{I}\left({x},{y}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{dilog}\left(\mathrm{1}−{xy}\right)−\mathrm{dilog}\left(\mathrm{1}+{xy}\right)\right) \\ $$
Commented by mohammad17 last updated on 16/Nov/20
whats the mean of dilog ?
$${whats}\:{the}\:{mean}\:{of}\:{dilog}\:? \\ $$
Commented by Olaf last updated on 17/Nov/20
The Spence function or dilogarithm, denoted Li_2 or dilog, is a special case of polylogarithm. Two special functions are called the Spence function : Li_2 (z) = −∫_0 ^z ((ln(1−u))/u)du = ∫_1 ^(1−z) ((lnt)/(1−t))dt, z∈C
$$\mathrm{The}\:\mathrm{Spence}\:\mathrm{function}\:\mathrm{or}\:\mathrm{dilogarithm}, \\ $$$$\mathrm{denoted}\:\mathrm{Li}_{\mathrm{2}} \:\mathrm{or}\:\mathrm{dilog},\:\mathrm{is}\:\mathrm{a}\:\mathrm{special}\:\mathrm{case}\:\mathrm{of} \\ $$$$\mathrm{polylogarithm}.\:\mathrm{Two}\:\mathrm{special}\:\mathrm{functions} \\ $$$$\mathrm{are}\:\mathrm{called}\:\mathrm{the}\:\mathrm{Spence}\:\mathrm{function}\:: \\ $$$$\mathrm{Li}_{\mathrm{2}} \left({z}\right)\:=\:−\int_{\mathrm{0}} ^{{z}} \frac{\mathrm{ln}\left(\mathrm{1}−{u}\right)}{{u}}{du}\:=\:\int_{\mathrm{1}} ^{\mathrm{1}−{z}} \frac{\mathrm{ln}{t}}{\mathrm{1}−{t}}{dt},\:{z}\in\mathbb{C} \\ $$

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