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I would like to know how this result from a textbook was obtained: $$ \int_0^1du\int_0^u\frac{u}{\sqrt{1-u}}\ln\frac{u}{v}\, dv = \int_0^1\frac{u^2}{\sqrt{1-u}}\, du=\frac{16}{15} $$

Is this some special result which I need to know? Otherwise if I simplify using the standard approach : $$ \int_0^1 \frac{u^2}{\sqrt{1-u}}\ln u\, du - \int_0^1\frac{u}{\sqrt{1-u}}\int_0^u (1)(\ln v)\,dv $$

which would require the value of $\ln(0)$.

I have verified that the result $\frac{16}{15}$ is indeed correct using WolframAlpha, but I do not know how the simplification from the text was obtained.

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  • $\begingroup$ $\int \frac{u^2}{\sqrt{1-u}} du = -\sqrt{1-u}(6u^2+8u+16)/15 $ $\endgroup$ – Ronald Blaak Nov 5 '18 at 14:47
  • $\begingroup$ @RonaldBlaak, thank you but I was really wondering about the first part and how to deal with the log function $\endgroup$ – NetUser5y62 Nov 5 '18 at 14:49
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This answer is trying to explain why $$\int_0^1du\int_0^u\frac{u}{\sqrt{1-u}}\ln\frac{u}{v}\, dv = \int_0^1\frac{u^2}{\sqrt{1-u}}\, du.$$


Integration by parts gives $$\int \ln(x) \,dx = x\ln(x) - x +C$$ for some constant $C.$ So we have $$\int_0^1 \ln(x)\,dx = [x\ln(x)]_0^1 - [x]_0^1 = [x\ln(x)]_0^1 -1.$$ Note that $$[x\ln(x)]_0^1 = - \lim_{t\to 0^+} t\ln(t) = -\lim_{t\to 0^+}\frac{\ln(t)}{\frac{1}{t}} = - \lim_{t\to 0^+}\frac{\frac{1}{t}}{-\frac{1}{t^2}} = \lim_{t\to 0^+} t = 0.$$ Therefore, $$\int_0^1 \ln(x)\,dx = [x\ln(x)]_0^1 -1 = -1.$$ Hence, $$\int_0^u \ln \left( \frac{u}{v} \right)\,dv = -\int_0^u \ln \left( \frac{v}{u} \right)\,dv = -u\int_0^1 \ln(x)\,dx = u$$ where I use substitution $x = \frac{v}{u}$ for second equality. The above implies that $$\int_0^1du\int_0^u\frac{u}{\sqrt{1-u}}\ln\left(\frac{u}{v}\right)\, dv = \int_0^1 \frac{u}{\sqrt{1-u}}\,du \int_0^u \ln\left( \frac{u}{v} \right)\,dv =\int_0^1\frac{u^2}{\sqrt{1-u}}\, du.$$

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If in $\int_{0}^{u}\frac{u}{\sqrt{1-u}}\log\frac{u}{v}\,dv $ we perform the substitution $v\mapsto u t$, we end up with $$ \int_{0}^{1}\frac{u^2(-\log t)}{\sqrt{1-u}}\,dt = \frac{u^2}{\sqrt{1-u}},$$ ($-\log t$ is integrable over $(0,1)$, even if it is unbounded) hence the original integral equals $$ \int_{0}^{1}\frac{u^2\,du}{\sqrt{1-u}}\stackrel{u\mapsto 1-u}{=}\int_{0}^{1}\frac{(1-u)^2}{\sqrt{u}}\,du \stackrel{u\mapsto u^2}{=}2\int_{0}^{1}(1-u^2)^2\,du,$$ reducing the original problem to the integration of a polynomial in $\mathbb{Z}[u]$, fairly straightforward.

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