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Question Number 201374 by MrGHK last updated on 05/Dec/23

	Answered by witcher3 last updated on 05/Dec/23

	$xy = z$ $\Rightarrow ydx = z$ $\int_{0}^{1} \int_{0}^{y} \frac{\tan^{- 1} (z) dzdy}{(1 + y) (1 + z)}$ $0 ⩽ z ⩽ y ⩽ 1 \Leftrightarrow 0 ⩽ z ⩽ 1 & z ⩽ y ⩽ 1$ $= \int_{0}^{1} \int_{z}^{1} \frac{\tan^{- 1} (z) dydz}{(1 + y) (1 + z)}$ $= \int_{0}^{1} \frac{\tan^{- 1} (z) \ln (2)}{1 + z} dz - \int_{0}^{1} \frac{\tan^{- 1} (z) \ln (1 + z)}{1 + z} dz = A - B$ $A easy$ $\tan^{- 1} (z) = \frac{1}{2 i} (\ln (1 - iz) - \ln (1 + iz))$ $= - IM \int_{0}^{1} \frac{\ln (1 + iz) \ln (1 + z)}{1 + z} dz$ $\int \frac{\ln (1 + iz)}{1 + z} dz, y = 1 + z$ $= \int \frac{\ln (1 + i - iy)}{y} dy$ $= \ln (1 + i) \ln (y) - {Li}_{2} (\frac{iy}{1 + i})$ $= \ln (1 + i) \ln (1 + z) - {Li}_{2} (\frac{i}{1 + i} (1 + z))$ $IBP \Rightarrow \int_{0}^{1} \frac{\ln (1 + iz) \ln (1 + z)}{1 + z} dz = Ω$ $= {[\ln (1 + i) \ln^{2} (1 + z) - \ln (1 + z) {Li}_{2} (\frac{i}{i + 1} (1 + z))]}_{0}^{1}$ $= \ln^{2} (2) \ln (1 + i) - \ln (2) {Li}_{2} (1 + i)$ $- \int_{0}^{1} \frac{\ln (1 + i) \ln (1 + z)}{1 + z} {dz}_{= \ln (1 + i) \frac{\ln^{2} (2)}{2}} + \int_{0}^{1} \frac{{Li}_{2} (\frac{i}{i + 1} (1 + z))}{1 + z} dz$ $\int \frac{{Li}_{2} (z)}{z} dz = {Li}_{3} (z)$ $Ω = \frac{\ln^{2} (2)}{2} \ln (1 + i) - \ln (2) {Li}_{2} (1 + i) + {Li}_{3} (i + 1) - {Li}_{3} (\frac{i}{i + 1})$ $we Can give closedForme$
	Commented by MrGHK last updated on 05/Dec/23

	$really nice but i used another approach$
	Commented by witcher3 last updated on 05/Dec/23

	$share it sir, when you have Times$
	Commented by MrGHK last updated on 05/Dec/23

	$h e r e i t i s$
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