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So the question is to evaluate the following definite integral

$$\int_0^u \frac{\ln (ax)}{\sqrt{u-x}} \rm dx $$

My approach is very close to get answer, but I couldn't finish.

I assumed $$f(a) =\int_0^u \frac{\ln (ax)}{\sqrt{u-x}} \rm dx $$ Now we've $$f'(a)= \int_0^u \frac{1}{ax} \cdot \frac{ x}{\sqrt{u-x}} \rm dx \implies f'(a) = \frac 1a \int_0^u \frac{1}{\sqrt{u-x}} \rm dx$$

$$f'(a) =-\frac{2 \sqrt{u-x}}{a} \Bigg |_0^u=\frac{2\sqrt u} a$$

$$\implies f(a) = \int \frac{2\sqrt u}{a} \rm da = 2\sqrt u (\ln (a)+C)$$

Here $C = g(u)$ is possible. But I can't find out a way to reach to the final answer.

Answer is $$\int_0^u \frac{\ln (ax)}{\sqrt{u-x}} \rm dx =2 \sqrt u ( \ln (4au)-2)$$

Can someone tell me how to proceed and/or can help evaluating this using any other method?

Thanks.

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Using integration by parts, with $f = \ln(x)$ and $dg = 1/\sqrt{1 - x}\, dx$, we have:

$$\int \frac{\ln(x)}{\sqrt{1 - x}}\, dx = -2\ln(x)\sqrt{1 - x} + 2\int \frac{\sqrt{1 - x}}{x}\, dx$$

Using the substitution $x = \sin^{2}(\theta)$, where $\theta \in (0,\pi/2)$, we get:

$$\int \frac{\sqrt{1 - x}}{x}\, dx = 2 \int \frac{\cos^{2}(\theta)}{\sin(\theta)}\, d\theta = 2\int \csc(\theta) - \sin(\theta)\, d\theta = -2\ln(\csc(\theta) + \cot(\theta)) + 2\cos(\theta) + C$$

So we have:

$$\int_0^{1}\frac{\ln(x)}{\sqrt{1 - x}}\, dx = \Big[-4\ln(\sin(\theta))\cos(\theta) - 4 \ln(\csc(\theta) + \cot(\theta)) + 4 \cos(\theta)\Big]_0^{\pi/2}$$

Writing $\ln(\csc(\theta) + \cot(\theta)) = \ln(1 + \cos(\theta)) - \ln(\sin(\theta))$, we get:

$$\int_0^{1}\frac{\ln(x)}{\sqrt{1 - x}}\, dx = 4\Big[\ln(\sin(\theta))(1 - \cos(\theta))\Big]_0^{\pi/2} + 4\Big[-\ln(1 + \cos(\theta)) + \cos(\theta)\Big]_0^{\pi/2}$$ $$= 4(\ln(2) - 1) + 4\lim_{\theta \rightarrow 0^{+}}\ln(\sin(\theta))(1 - \cos(\theta))$$ You can show that this limit equals zero, for example by multiplying by $(1 + \cos(\theta))/(1 + \cos(\theta))$ and using the fact that $x\ln(x) \rightarrow 0$ as $x \rightarrow 0$.

Finally, making the substitution $x = t/u$ where $u$ is a constant:

$$\int_0^{1}\frac{\ln(x)}{\sqrt{1 - x}}\, dx = \int_0^{u}\frac{\ln(t/u)}{\sqrt{1 - t/u}}\frac{\, dt}{u} = \frac{1}{\sqrt{u}}\int_0^{u}\frac{\ln(t/u)}{\sqrt{u - t}}\, dt = \frac{1}{\sqrt{u}}f(1/u)$$.

Where $f$ is the function you've defined, i.e. $f(a) = 2\sqrt{u}(\ln(a) + C)$. Putting it all together:

$$4(\ln(2) - 1) = 2(\ln(1/u) + C) \Rightarrow C = 2\ln(2) - \ln(1/u) - 2 = \ln(4u) - 2$$

So $f(a) = 2\sqrt{u}(\ln(a) + \ln(4u) - 2) = 2\sqrt{u}(\ln(4au) - 2)$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{u}{\ln\pars{ax} \over \root{u - x}}\,\dd x & = \int_{0}^{u}{\ln\pars{au\bracks{x/u}} \over \root{u}\root{1 - x/u}}\,u \,{\dd x \over u} = \root{u}\int_{0}^{1}\ln\pars{aux}\pars{1 - x}^{-1/2}\,\dd x \\[5mm] & = \root{u}\,\left.\partiald{}{z}\int_{0}^{1}\pars{aux}^{\,\large z} \,\pars{1 - x}^{-1/2}\,\dd x\,\right\vert_{\ z\ =\ 0} \\[5mm] & = \root{u}\,\partiald{}{z}\bracks{\pars{au}^{\large\, z}\int_{0}^{1}x^{\,\large z} \,\pars{1 - x}^{-1/2}\,\dd x}_{\ z\ =\ 0} \\[5mm] & = \root{u}\,\partiald{}{z}\bracks{\pars{au}^{\large\, z}\, {\Gamma\pars{z + 1}\Gamma\pars{1/2} \over \Gamma\pars{z + 3/2}}}_{\ z\ =\ 0} = \bbx{2\root{u}\bracks{\ln\pars{4au} - 2}} \end{align}

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Both the $a$ and $u$-parameters are quite irrelevant, the actual question is just about $\int_{0}^{1}\frac{\log x}{\sqrt{1-x}}\,dx$, which can be computed by enforcing the substitution $x=\sin^2\theta$ and exploiting the Fourier series of $\log\sin$, for instance. Alternative approaches involve the differentiation of a Beta function or the Fourier-Legendre expansions of $\log(x)$ and $\frac{1}{\sqrt{1-x}}$, which are pretty simple to compute from Rodrigues' formula:

$$ \log(x) = -1+\sum_{n\geq 1}(-1)^{n+1}\left(\frac{1}{n}+\frac{1}{n+1}\right)P_n(2x-1) $$ $$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0} 2 P_n(2x-1) $$ $$ \int_{0}^{1}\frac{\log x}{\sqrt{1-x}}\,dx = 4(\log 2-1).$$

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  • $\begingroup$ Thanks for your answer, but I haven't studied so much Higher Mathematics so far. I am a highschool student. (Although I am not sure highschool Math is sufficient to evaluate this). Can we continue my method? Is it possible to reach to the answer using that? $\endgroup$ – Jaideep Khare Feb 26 '18 at 19:10

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