Evaluating $\int_{0}^{1} \frac{\ln(x)}{\sqrt{1-x}} dx$ using different methods

Question

I was evaluating $\displaystyle\int_0^1\frac{\ln(x)}{\sqrt{1-x}} dx$ using different method, given below. First I used $u = 1-x$ to get $\displaystyle\int_0^1\frac{\ln(1-x)}{\sqrt{x}} dx$

Try 1

Easily we can use parts integration to get:

$$I = 2 \sqrt{x} \ln(1-x) \biggr|_{0}^{1} + \int_{0}^{1} \frac{2\sqrt{x}}{1-x} dx$$

For last integral, I used $x = t^2$ to get:

$$\begin{align} \int_{0}^{1} \frac{2\sqrt{x}}{1-x} dx &= \frac{4t^2}{1-t^2} dt\\ & = 4 \int_0^1\left(-1+\frac{1}{1-t^2}\right) dt\\ & = -4t + 2 \ln\left(\frac{1+t}{1-t}\right) \end{align}$$

So we get

$$I = 2 \sqrt{x} \ln(1-x) - 4 \sqrt{x} + 2 \ln\left(\frac{1+\sqrt x}{1-\sqrt x}\right) \rvert_{0}^{1}\\ = \left.2\ln\left(\frac{(1-x)^{\sqrt x} (1+\sqrt{x})}{(1-\sqrt{x})}\right) \right|_0^1$$

which upon taking limits we get $4\ln(2)-4$.

Try 2

I tried series of $\ln(1-x)$ as its in domain.

$$I = -\int_0^1 \sum_{i=1}^{\infty} \frac{x^{\frac{2i-1}{2}}}{i} \\ = -2\sum_{1}^{\infty} \frac{1}{i(2i+1)} \\ = -2 \sum_{i=1}^\infty\left(\frac{1}{i}-\frac{2}{2i+1}\right)$$

But now I rearranged terms (illegal) to get $2 \ln(2)-4$ which my miss said is not correct. I am not able to use series here, how to correct this error that is without rearranging?

Also please supply more ways of solving. Note i am not in higher class so please dont expect me to understand complicated maths.

Answer 1

Let $$ f(x)=\sum_{1}^{\infty} \frac{1}{i(2i+1)}x^{2i+1}$$ and then $$ f(0)=0,f'(0)=0, f''(x)=2\sum_{1}^{\infty}x^{2i-1}=\frac{2x}{1-x^2}. $$ So $$ f(1)=\int_0^1\int_0^t\frac{2x}{1-x^2}dxdt\\ =\int_0^1\int_x^1\frac{2x}{1-x^2}dtdx\\ =2\int_0^1\frac{x}{1+x}dx=2(1-\ln2) $$ and hence $$I = -2\sum_{1}^{\infty} \frac{1}{i(2i+1)} \\ =-4(1-\ln2).$$

Answer 2

Another way (I don't know if it is too complicated or not, please tell me if something is not clear). Recalling the definition of the Beta function we consider $$B\left(a,1/2\right)=\int_{0}^{1}\frac{x^{a-1}}{\sqrt{1-x}}dx.$$ Now by the Leibniz integral rule we have $$\frac{\partial}{\partial a}B\left(a,1/2\right)=\int_{0}^{1}x^{a-1}\frac{\log\left(x\right)}{\sqrt{1-x}}dx.$$ But $$\frac{\partial}{\partial a}B\left(a,1/2\right)=B\left(a,1/2\right)\left(\psi\left(a\right)-\psi\left(a+\frac{1}{2}\right)\right)$$ where $\psi\left(x\right)$ is the Digamma function. This identity follows from the representation$$B\left(x,y\right)=\frac{\Gamma\left(x\right)\Gamma\left(y\right)}{\Gamma\left(x+y\right)},$$ where $\Gamma\left(x\right)$ is the Gamma function. So taking $a=1$ we get $$\int_{0}^{1}\frac{\log\left(x\right)}{\sqrt{1-x}}dx=B\left(1,1/2\right)\left(\psi\left(1\right)-\psi\left(\frac{3}{2}\right)\right)=\color{red}{\log\left(16\right)-4}.$$ Note that it is possible to calculate explicitily these functions with these particular values: see for example this and this.