Does the following integral converge or diverge for $t\rightarrow \infty$?
$$\int_1^t \frac{1}{2}\cdot\frac{1+\cos x}{x}dx $$
If anyone wants to know what I've tried so far, sorry, but I haven't looked at this sort of problem in years. I'm pretty rusty. The first term is obviously divergent. I can tell that much. I am wondering if the negative contributions from the $\cos x$ term are enough to prevent an overall divergence.
On
DIVERGENT. HINT: For each $n \in \mathbb{N}$:
$$\int_{2\pi n}^{2\pi(n+1)} \frac{1+\cos x}{x} dx$$ $$\ge \ \frac{1}{2\pi(n+1)}\int_0^{2\pi} (\cos x+1)dx$$ $$\ge \ \frac{1}{2\pi (n+1)}×\frac{1}{4}.$$
[Indeed, $\cos x +1$ is nonnegative for all $x$ and so $\frac{1+\cos x}{dx}$ is lower-bounded by $\frac{1+\cos x}{2\pi(n+1)}$ for all $x\in [2\pi n, 2\pi(n+1)]$. Finally, use
$$\int_0^{2\pi} (1+\cos x)dx \ \ge \ \frac{1}{4}.]$$
Can you use this to conclude for each $n \in \mathbb{N}$ the inequality $$\int_{2\pi}^{2\pi(n+1)} \frac{1+\cos x}{x}dx$$ $$\ge \ \frac{1}{4}×\sum_{i=1}^{n} \frac{1}{2\pi(n+1)}.$$
On
We have $${1+\cos x\over 2}=\cos^2(x/2)=\sin^2((x+\pi)/2)$$ Thus for $t>1+\pi$ we get $$\int\limits_1^t{1+\cos x\over x}\,dx=\int\limits_1^t{\cos^2(x/2)\over x}dx+\int\limits_1^t{\sin^2((x+\pi)/2)\over x}dx\\ =\int\limits_1^t{\cos^2(x/2)\over x}dx+\int\limits_{1+\pi}^{t+\pi}{\sin^2(x/2)\over x}dx \ge \int\limits_{1+\pi}^t{dx\over x} $$ Remark The reasoning above can be used for proving that the integral $$\int\limits_1^\infty {a+\cos x\over x}\,dx$$ is divergent for $a\neq 0.$ Indeed for $a>0$ we have $$\int\limits_1^t{a+\cos x\over x}\,dx= (a-1) \int\limits_1^t{dx\over x}+\int\limits_1^t{1+\cos x\over x}\,dx\\ \ge (a-1) \int\limits_1^t{dx\over x}+ \int\limits_{1+\pi}^t{dx\over x} \\ = a \int\limits_{1+\pi}^t{dx\over x} +(a-1)\int\limits_1^{1+\pi}{dx\over x}$$
It diverges to $\infty$. One of the things you should always try when dealing with oscillatory integrals is integration by parts. Let us ignore the factor of $\frac{1}{2}$. Here, we have \begin{align} \int_1^t\frac{1+\cos x}{x}\,dx&=\log t+\int_1^t\frac{\cos x}{x}\,dx\\ &=\log t+\left[\frac{\sin x}{x}\right]_1^t-\int_1^t\left(-\frac{1}{x^2}\right)\sin x\,dx\\ &=\log t+\left[\frac{\sin t}{t}-\sin (1)\right]+\int_1^t\frac{\sin x}{x^2}\,dx. \end{align} As $t\to\infty$, we have $\log t\to\infty$, and $\frac{\sin t}{t}\to 0$ and the last integral converges absolutely due to the $\frac{1}{x^2}$ decay. So, putting it all together, the LHS diverges to $\infty$.
So, the main culprit here is the $\frac{1}{x}$ term because that gives a $\log t$ divergence. The $\frac{\cos x}{x}$ term has sufficient oscillation (and an integration by parts reveals this) that its improper integral as $t\to\infty$ converges.