Subject: Mathematics — Definite Integration
Question: Let $\alpha$ and $\beta$ be the real numbers such that
$$\lim_{x \rightarrow 0}\frac{\frac{\alpha}{2}\left( \int_{1}^{x}\frac{1}{1 - t^{2}}\, dt \right) + \beta x\cos x}{x^{3}} = 2.$$
Then the value of $\alpha + \beta$ is \_\_\_\_\_\_\_\_\_\_.
Answer: $2.4$
Solution: $\lim_{x \rightarrow 0}\frac{\frac{\alpha}{2}\left( \int_{1}^{x}\frac{1}{1 - t^{2}}\, dt \right) + \beta x\cos x}{x^{3}} = 2.$
Applying L' Hospital's rule,
$$\lim_{x \rightarrow 0}\frac{\frac{\alpha}{2}\left( \frac{1}{1 - x^{2}} \right) + \beta\cos x - \beta x\sin x}{3x^{2}} = 2.$$
$$\Rightarrow \lim_{x \rightarrow 0}\frac{\frac{\alpha}{2} + \beta\cos x\left( 1 - x^{2} \right) - \beta x\sin x\left( 1 - x^{2} \right)}{3x^{2}\left( 1 - x^{2} \right)} = 2.$$
For the limit to exist,
$$\frac{\alpha}{2} + \beta = 0\quad\quad\cdots(i)$$
$$\Rightarrow \lim_{x \rightarrow 0}\frac{- \beta + \beta\left( 1 - x^{2} \right)\left( 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \cdots \right) - \beta x\left( 1 - x^{2} \right)\left( x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots \right)}{3x^{2} \times 1} = 2.$$
$$\Rightarrow \lim_{x \rightarrow 0}\frac{\frac{\beta}{3}\left\{ - 1 + \left( 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \cdots \right) + \left( - x^{2} + \frac{x^{4}}{2!} - \frac{x^{6}}{4!} + \cdots \right) + \left( - x^{2} + \frac{x^{4}}{3!} - \frac{x^{6}}{5!} + \cdots \right) + \left( x^{4} - \frac{x^{6}}{3!} + \frac{x^{8}}{5!} - \cdots \right) \right\}}{x^{2}}$$
$$= 2.$$
$$\Rightarrow \frac{\beta}{3}\left( - \frac{1}{2} - 1 - 1 \right) = 2$$
$$\Rightarrow \beta = - \frac{12}{5},\quad\quad\alpha = \frac{24}{5}.$$
$$\alpha + \beta = \frac{12}{5} = 2.4.$$
##