Summation Contest Conceptual Problem-2

We have $\begin{aligned} \int_0^1 \frac{x(1-x)^{\frac14}}{(2-x)^2}\,dx & = \int_0^1 \frac{x^{\frac14}(1-x)}{(1+x)^2}\,dx \\ & = 2\int_0^1 \frac{x^{\frac14}}{(1+x)^2}\,dx - \int_0^1 x^{-\frac34}\,dx + \int_0^1 \frac{x^{-\frac34}}{1 + x},dx \\ & = 2B_{\frac12}\big(\tfrac54,\tfrac34\big) - 4 + B_{\frac12}\big(\tfrac14,\tfrac34\big) \\ & = -5 + \tfrac34\pi\sqrt{2} + \tfrac32\sqrt{2}\ln(\sqrt{2}+1) \end{aligned}$ making the answer $5+3+2+4+3+2+2+2 = \boxed{23}$ .

I want to convert the integral into a summation and then convert it back into an integral which would be easier to solve.(Maybe a direct transformation exist but right now it is not coming to my mind) Let us re-write the integration as:-

$\displaystyle \int_{0}^{1}\frac{(1-x)x^{\frac{1}{4}}}{(1+x)^{2}} dx \quad\quad\quad \text{(using substitution x=1-u)}$

Now we know :-

$\displaystyle \frac{1}{1+x} = \sum_{r=0}^{\infty}(-1)^{r}x^{r},$ ( $-1<x<1$ )

Differentiating both sides :-

$\displaystyle \frac{1}{(1+x)^{2}}=\sum_{r=1}^{\infty}(-1)^{r-1}rx^{r-1}$

So we have :-

$\displaystyle \sum_{r=1}^{\infty} \int_{0}^{1}(-1)^{r-1}rx^{r-\frac{3}{4}}(1-x)dx$

$\displaystyle \sum_{r=1}^{\infty} (-1)^{r-1} \left(\frac{4r}{4r+1} -\frac{4r}{4r+5}\right)$

$\displaystyle \sum_{r=1}^{\infty} (-1)^{r-1}\frac{5}{4r+5} - \sum_{r=1}^{\infty}(-1)^{r-1}\frac{1}{4r+1} \quad\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \text{(both series as well as original series are convergent by Leibnitz test})$

Now using $\displaystyle \frac{1}{1+x} = \sum_{r=0}^{\infty}(-1)^{r}x^{r}$ we have:-

$\displaystyle \frac{1}{1+x^{4}} = \sum_{r=0}^{\infty}(-1)^{r}x^{4r}$

Hence $\displaystyle \frac{x^{4}}{1+x^{4}} = \sum_{r=1}^{\infty}(-1)^{r-1}x^{4r}$ .

Hence $\displaystyle \int_{0}^{1}\frac{x^{4}}{1+x^{4}}dx = \sum_{r=1}^{\infty}(-1)^{r-1}\frac{1}{4r+1}$ .

Similarly $\displaystyle \int_{0}^{1}\frac{x^{8}}{1+x^{4}}dx = \sum_{r=1}^{\infty}(-1)^{r-1}\frac{1}{4r+5}$ .

Hence we have our transformed integral as :- $\int_{0}^{1} \frac{5x^{8}-x^{4}}{1+x^{4}}dx$

$\displaystyle = 5\int_{0}^{1}x^{4} dx - 6\int_{0}^{1}\frac{x^{4}}{1+x^{4}}dx = 1-6\int_{0}^{1}\left(\frac{x^{4}+1}{x^{4}+1} -\frac{1}{1+x^{4}}\right)dx$

$\displaystyle = -5 +6\int_{0}^{1} \frac{1}{1+x^{4}} dx$

Now the function $\displaystyle \frac{1}{1+x^{4}}$ has a known antiderivative and it's indefinite integration is taught in high school.

$\displaystyle \int\frac{1}{1+x^{4}}dx = \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right) - \frac{1}{4\sqrt{2}}\ln\left(\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right) + C$

Using the above we arrive at the answer: -

$\displaystyle -5 + \frac{3\pi\sqrt{2}}{4} +\frac{3\sqrt{2}}{2}\ln(\sqrt{2}+1)$