We want to evaluate $\lim_{x \to 0} \int_0^x \frac{\cos(t^3)}{t+x}dt$.
I guess the answer is $\log2$. But I'm not sure about correctness of my ideas. Let me show them :
My attempts :
1) $\displaystyle \lim_{x \to 0} \int_0^x \frac{\cos(t^3)}{t+x}dt = \lim_{x \to 0} \lim_{n \to \infty} \sum_{k = 0}^{n} \frac{\cos(\left(\frac{xk}{n}\right)^3)}{n + \frac{xk}{n}} \cdot \frac{x}{n} = \lim_{x \to 0} \lim_{n \to \infty}\sum_{k = 0}^{n} \frac{\cos(\left(\frac{xk}{n}\right)^3)}{k + n} = \lim_{n \to \infty}\sum_{k = 0}^{n} \frac{1}{k+n} = \log{2}$
But I'm not sure about correctness of changing the limits.
2)$\displaystyle \int_0^x\frac{\cos(t^3)}{t+x}dt = \cos(x^3)\log{2x} - \log{x} - \int_0^x 3t^2\log{(t+x)} \sin{t^3} dt$. First two terms in limit gives us $\log{2}$ and the last integral part should be zero in limit?
If we put $$t=xu$$
the integral becomes
$$\int_0^1\frac{\cos(x^3u^3)}{1+u}du$$ The integrand, as a two variables $(x,u)$ function is continuous at $[-1,1]\times [0,1]$, the limit when $x\to 0 $ will be
$$\int_0^1\frac{\cos(0)du}{1+u}=\int_0^1\frac{du}{1+u}=\ln(2)$$