Showing that $\int_{1}^{\infty}\cos(x^2)\arctan(x)dx$ converges Hey I wanted to show that
$$\int_{1}^{\infty}\cos(x^2)\arctan(x)dx$$ converges (at least that's what Wolfram Alpha says).
I know that
$$\int_{1}^{\infty}\frac{\sin(x)}{x}\arctan(x)dx$$ converges.
 A: hint
The idea is to write the integrand as a sum of three functions.
You might know that
$$(\forall x>0)\;\;$$
$$ \arctan(x)=\frac{\pi}{2}-\arctan(\frac 1x)$$
By the substitution $ x^2=t $, we prove that $$\frac{\pi}{2}\int_1^{+\infty}\cos(x^2)dx$$ which becomes
$$\int_1^{+\infty}\frac{\cos(t)dt}{2\sqrt{t}}\text{ is convergent}$$
for the second part, near $ +\infty, $
$$\arctan(\frac 1x)=\frac 1x-\frac{1}{3x^3}(1+\epsilon(x))$$
$$=f_1(x)+f_2(x)$$
with
$$\int_1^{+\infty}\cos(x^2)f_1(x)dx$$
convergent.
finally
$$|\cos(x^2)f_2(x)|<\frac{2}{3x^3}$$
So, $$\int_1^{+\infty}\cos(x^2)f_2(x)dx$$ is convergent by comparison test.
we conclude that your integrale is convergent as a sum of three convergent integrales.
A: Let $f(x) = \frac{\arctan x}{x}$. Notice $f'(x) = -\frac1x\left(f(x) - \frac{1}{1+x^2}\right)$.
For $x > 0$, apply MVT  to $f(x)$ over interval $[0,x]$, one can find a $y$ in $(0,x)$ such that
$$f(x) = \frac{\arctan x}{x} = \frac{1}{1+y^2} > \frac{1}{1+x^2} > 0$$
This implies $f'(x) < 0$ and hence $f(x)$ is positive decreasing for $x > 0$.
Rewrite the integral at hand as
$$\int_1^\infty \cos(x^2)\arctan(x) dx
= \lim_{p\to\infty} \int_1^p \left(\frac{\arctan x}{x}\right) x\cos(x^2) dx$$
Since
$$\left|\int_1^p x\cos(x^2) dx\right| = 
\frac12\left|\sin p^2 - \sin 1\right| \le \frac12\left|1+\sin 1\right|$$
is bounded from above by a number independent of $p$.
By a variant of Dirichlet's test for improper integrals, your integral converges.
A: With the help of your ideas and a little bit more of studying, I developed an own solution and approach with some more criteria and a little bit more technically:
With the substitution $y=u(x)=x^{2}$ we get
$$
\begin{aligned}
\int_{1}^{+\infty} \cos \left(x^{2}\right) \arctan(x) d x &=\int_{u(1)}^{u(+\infty)} \cos(y) \arctan(\sqrt{y}) \frac{d y}{2 \sqrt{y}} \\
&=\frac{1}{2} \int_{1}^{+\infty} \frac{\cos(y)}{\sqrt{y}} \arctan (\sqrt{y}) d y .
\end{aligned}
$$
Let us first prove that the integral
$$
\int_{1}^{+\infty} \frac{\cos(y)}{\sqrt{y}} d y
$$
converges. Let us set $f(y)=\cos(y)$ and $g(y)=\frac{1}{\sqrt{y}}$. Since the function
$$
\int_{1}^{x} f(y) d y=\int_{1}^{x} \cos(y) d y=\sin(x)-\sin(1)
$$
is bounded and $g(y)$ is monotonic and $g(x) \rightarrow 0$ for $x \rightarrow \infty$, then by the Dirichlet criterion we get that the integral $\int_{1}^{+\infty} \frac{\cos(y)}{\sqrt{y}} d y$ converges.
Let us now set $f(y)=\frac{\cos (y)}{\sqrt{y}}$ and $g(y)=\arctan (\sqrt{y}) .$ Since $\int_{1}^{\infty} f(y) d y$ is convergent and $g$
is monotone and bounded, the integral
$$
\int_{1}^{+\infty} \frac{\cos(y)}{\sqrt{y}} \arctan (\sqrt{y}) d y
$$
is also converging according to the Abel criterion. Thus the integral
$$
\int_{1}^{+\infty} \cos \left(x^{2}\right) \arctan(x) d x
$$
also converges.
Let us prove that this integral is conditionally convergent, i.e.
$$
\int_{1}^{+\infty}\left|\cos \left(x^{2}\right) \arctan(x)\right| d x=+\infty
$$
Since for $x>1$ holds.
$$
\arctan(x) \geq \arctan (1)=\frac{\pi}{4}
$$
it is sufficient to prove that
\begin{equation}
    \int_{1}^{+\infty}\left|\cos \left(x^{2}\right)\right| d x=+\infty .
\end{equation}
We have as before
$$
\int_{1}^{+\infty}\left|\cos \left(x^{2}\right)\right| d x=\frac{1}{2} \int_{1}^{+\infty} \frac{|\cos(y)|}{\sqrt{y}} d y .
$$
The zeros of $\cos(y)$ in the interval $[1, \infty)$ are $y_{n}=\pi / 2+\pi n, n=0,1,2 .... .$Since in each interval$\left(y_{n-1}, y_{n}\right)$ the function $\cos(y)$ does not change sign, we obtain
\begin{aligned}
\int_{y_{n-1}}^{y_n}\frac{|\cos(y)|}{\sqrt{y}}dy &\geq\frac{1}{\sqrt{y_n}} \int_{y_{n-1}}^{y_n}|\cos(y)| d y \\
&=\frac{1}{\sqrt{y_n}} \left|\int_{y_{n-1}}^{y_n} \cos(y) dy \right| \\
&=\frac{1}{\sqrt{y_n}}\left|[\sin(y)]_{y_{n-1}}^{y_n}\right| =\frac{2}{\sqrt{y_n}}
\end{aligned}
from which follows
\begin{aligned}
\int_{1}^{+\infty}\left|\cos \left(x^{2}\right)\right| d x &=\frac{1}{2} \int_{1}^{+\infty} \frac{|\cos(y)|}{\sqrt{y}} d y \\
&\geq \frac{1}{2} \sum_{y=1}^{\infty} \int_{y_{n-1}}^{y_n} \frac{|\cos(y)|}{\sqrt{y}} d y \\
&\geq \sum_{n=1}^{\infty} \frac{1}{\sqrt{y_n}}
\end{aligned}
Since $y_{n} \sim \pi n$ and $\sqrt{y_{n}} \sim \sqrt{\pi n}$ for $n \rightarrow \infty$ and $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}+\infty,$$ we obtain
$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{y_n}}=\infty$$
from which (1) follows.
