The theoretical tools at my disposal are Abel's test and Dirichlet's test. To recap those: Say I have an integral of the form $$\int_{a}^{b}f\cdot g \hspace{1.5mm} dx$$ with Improperness (vertical or horizontal asymptote) at b.

Abel's test guarantees convergence for

$\bullet$ $g$ monotone and bounded on $(a,b)$ $\hspace{5mm}$ $\bullet$ $\int_{a}^{b}f $ convergent.

Dirichlet's test guarantees convergence for

$\bullet$ $g$ monotone on $(a,b)$ and $lim_{x\to b}\hspace{1.5mm} g(x) = 0 $ $\hspace{5mm}$ $\bullet$ $\lim_{\beta \to b}$ $\int_{a}^{\beta}f $ bounded.

So for $\int_{1}^{\infty} \frac{cos(x)}{x^r} dx$, $cos(x)$ must be my $f$ and since $\int_{1}^{\infty} cosx \hspace{1mm} dx$ diverges, I must show it is bounded, which is intuitively easy given the graph of $cos(x). $ But how to make it precise? With that done, for $r > 0 $ I get my bounded function $ g(x)=\frac{1}{x^r} $ as required by Dirichlet's test, right?

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    $\begingroup$ i think $r>0$ is right $\endgroup$ – Dr. Sonnhard Graubner May 6 '17 at 8:20
  • $\begingroup$ The convergence will be conditional for $0<r<1$, not absolute right? Since I need the negative "Areas" of the cosine function for convergence. But for $r >1$ I get absolute convergence, correct? $\endgroup$ – ghthorpe May 6 '17 at 8:26
  • $\begingroup$ @ghthorpe you left out $r=1$. $\endgroup$ – Oscar Lanzi May 6 '17 at 9:27
  • $\begingroup$ $\int_1^\infty\cos x\mathrm dx$ is bounded because $\int_0^y\cos x\mathrm dx=\sin y$ what is bounded. $\endgroup$ – Masacroso May 6 '17 at 9:29
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    $\begingroup$ Integrating by parts $\displaystyle \int_1^\infty \cos(x)x^{-r}dx=\lim_{t \to \infty}\int_1^{t} \cos(x)x^{-r}dx=\lim_{t \to \infty} \sin(r)x^{-r}|_1^{t} +\int_1^{t} \sin(x) r x^{-r-1}dx$ converges for $r > 0$, but $\displaystyle\lim_{n \to \infty, n \in \mathbb{Z}}\int_1^{2\pi n} \cos(x)x^{-r}dx$ converges for $r > -1$ $\endgroup$ – reuns May 6 '17 at 10:29

Disclaimer: This is my first answer. And my English is very bad.

First, you know the inequality $$(\forall r \in \mathbb{R})~~~ \frac{\cos(x)}{x^r} \leq \frac{1}{x^r}$$

With integration by parts you can prove convergence for $r=1$ using $$\frac{\cos(x)}{x} \leq \frac{1}{x}$$

For $r > 1$ again use integration by parts: $\int_{1}^{+\infty} \frac{\cos(x)}{x^a} dx= \left[\frac{\sin(x)}{x^r}\right]_1^{+\infty} + r\int_1^{+\infty} \frac{\cos(x)}{x^{r+1}}dx \leq \cos (1)+ r\int_1^{+\infty} \frac{\cos(x)}{x^{r+1}} dx$ and

$$\int_1^{+\infty} \frac{|\cos(x)|}{x^{r+1}} dx\leq \int_1^{+\infty} \frac{1}{x^{r+1}} dx<\infty.$$ Hence your integral converges on $[1,+\infty) \Leftrightarrow r>1$.

Similar arguments show divergence for $r < 1$.

  • $\begingroup$ but the integral converges for $r>0$. $\endgroup$ – Masacroso May 6 '17 at 22:49

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