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 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

$$\mathrm{The}\:\mathrm{Spence}\:\mathrm{function}\:\mathrm{or}\:\mathrm{dilogarithm}, \\ $$$$\mathrm{denoted}\:\mathrm{Li}_{\mathrm{2}} \:\mathrm{or}\:\mathrm{dilog},\:\mathrm{is}\:\mathrm{a}\:\mathrm{special}\:\mathrm{case}\:\mathrm{of} \\ $$$$\mathrm{polylogarithm}.\:\mathrm{Two}\:\mathrm{special}\:\mathrm{functions} \\ $$$$\mathrm{are}\:\mathrm{called}\:\mathrm{the}\:\mathrm{Spence}\:\mathrm{function}\:: \\ $$$$\mathrm{Li}_{\mathrm{2}} \left({z}\right)\:=\:−\int_{\mathrm{0}} ^{{z}} \frac{\mathrm{ln}\left(\mathrm{1}−{u}\right)}{{u}}{du}\:=\:\int_{\mathrm{1}} ^{\mathrm{1}−{z}} \frac{\mathrm{ln}{t}}{\mathrm{1}−{t}}{dt},\:{z}\in\mathbb{C} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com