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.
Best Answer
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.