Evaluating $\lim_{x\to0^+}x\int_x^1\frac{\cos t}{t^2}dt$ [duplicate]

Find $$\lim_{x \to 0^+}x\int_x^1 \dfrac{\cos t}{t^2}\,dt$$

This looks like an interesting problem,but i cannot figure out where to start, can anyone explain

Solution 1:

$$\cos(t)=1+O(t^2)$$ $$\frac{\cos(t)}{t^2}=\frac{1}{t^2}+O(1)$$ $$\int_{x}^1\frac{\cos(t)}{t^2}dt=-1+\frac{1}{x}+O(1)$$ $$x\int_{x}^1\frac{\cos(t)}{t^2}dt=-x+1+O(x)$$ $$\lim\limits_{x \to 0^+}x\int_x^1 \dfrac{\cos t}{t^2}\hspace{1mm}dt=1$$

Solution 2:

You may start like this :

$$\lim\limits_{x \to 0^+}x\int_x^1 \dfrac{\cos t}{t^2}\hspace{1mm}dt = \lim\limits_{x \to 0^+}\dfrac{\int_x^1 \dfrac{\cos t}{t^2}\hspace{1mm}dt}{1/x}$$

and see if it satisfies the hypothesis for applying L'Hopital's rule

Solution 3:

Substituting $s = \frac{1}{t}$ and $y = \frac{1}{x}$ you get $$\lim_{x \rightarrow 0^+} x \int_x^1 \frac{\cos t}{t^2}dt = \lim_{x \rightarrow 0^+} x\int_1^{\frac{1}{x}} \cos \left( \frac{1}{s} \right) ds = \lim_{y \rightarrow + \infty} \frac{\int_1^y \cos \left( \frac{1}{s} \right) ds}{y} $$

Now, applying De L'Hopital we get

$$\lim_{y \rightarrow + \infty} \frac{\int_1^y \cos \left( \frac{1}{s} \right) ds}{y}= \lim_{y \rightarrow + \infty} \frac{\cos \left( \frac{1}{y} \right)}{1} = \cos0 = 1 $$

Solution 4:

Substituting $u= \dfrac{t}{x}$, one has $$x\int_x^1\frac{\cos t}{t^2}dt = \int_{1}^{+\infty} \frac{\cos(ux)}{u^2} \chi_{\lbrace xu \leq 1\rbrace} du$$

Hence, Lebesgue's dominated convergence gives directly that

$$\lim_{x \rightarrow 0^+} x\int_x^1\frac{\cos t}{t^2}dt = \int_{1}^{+\infty} \frac{1}{u^2} du$$

i.e. that $$\boxed{\lim_{x \rightarrow 0^+} x\int_x^1\frac{\cos t}{t^2}dt=1}$$