Study of a function and other facts Study the function $f:(0,+\infty) \rightarrow \mathbb{R}$ defined by: $$f(x) = \int_0^{+\infty} \arctan(t/x)e^{-t}dt \qquad (x>0)$$


*

*Prove that $f$ is continuous, monotone and convex.

*Evaluate:
$$\lim_{x \rightarrow 0}f(x), \qquad \lim_{x \rightarrow +\infty}f(x), \qquad \sup_{x>0} f(x), \qquad \inf_{x>0} f(x).$$

*Show that $f$ is differentiable and find $f'$.
My try:


*

*I'm not sure about this, but I think that if I show that: $$f(x) = \int_0^{+\infty} \arctan(t/x)e^{-t}dt$$ is bounded and the integrating is continuous I've done. Easy to see that: $$0<\int_0^{+\infty} \arctan(t/x)e^{-t}dt<\frac{\pi}{2}e^{-t}$$ and the integrating is continuous for each $x>0$. Moreover to prove that $f$ is monotone I have to show that $f(x+1)<f(x)$ for all $x>0$ that could be easy shown by the fact that $\arctan(x)$ is strictly increasing. About the convexity I should prove first that $f$ is two times differentiable… maybe it's easier to use the definition?

 A: *

*
*

*The function \begin{array}{ccccc}
K & : & (0,\infty)\times \mathbb R^+ & \to & \mathbb R \\
 & & (x,t) & \mapsto & \arctan(t/x)e^{-t} \\
\end{array}



is clearly continuous.
Furthermore, $\forall (x,t)\in (0,\infty)\times \mathbb R^+,|K(x,t)|\leq \frac{\pi}{2}e^{-t}$
Since the function $t\to \frac{\pi}{2}e^{-t}$ is integrable over $\mathbb R^+$, the previous inequality grants domination.
You can then deduce that $f$ is continuous.


*

*Let $x,y\in (0,\infty)$ such that $x<y$ and $t\in \mathbb R^+$


Then, $1/y <1/x \Rightarrow \arctan(t/y)e^{-t}< \arctan(t/x)e^{-t}$
Integrating the inequality with respect to $t$ yields $f(y)< f(x)$
$f$ is consequently decreasing.


*

*Fix some arbitrary non-negative $t$.


We need to prove the function $g_t:\displaystyle x\to\arctan(\frac{t}{x})e^{-t}$ is convex
The derivative of $g_t$ is $\displaystyle g_t'(x)=\frac{-te^{-t}}{t^2+x^2}$ which an increasing function. Hence the convexity of $g$.
The following inequality holds $\forall \lambda \in [0,1], \forall (x,y)\in (0,\infty), g_t((1-\lambda)x+\lambda y)\leq (1-\lambda)g_t(x)+\lambda g_t(y) \; \; \; \; (1)$
Now, $\displaystyle \forall \lambda \in [0,1], \forall (x,y)\in (0,\infty), f((1-\lambda)x+\lambda y) = \int_0^{\infty} g_t((1-\lambda)x+\lambda y) dt \; \; \; \; (2)$
$(1)$ and $(2)$ imply $\displaystyle \forall \lambda \in [0,1], \forall (x,y)\in (0,\infty), f((1-\lambda)x+\lambda y) \leq (1-\lambda)f(x)+\lambda f(y) $
Hence the convexity of $f$.



*

*I'm going to do it the hard way (by means of sequences)


Let $(x_n)$ be an arbitrary sequence of positive numbers that goes to $0$
Let us prove that $f(x_n)$ goes to $\displaystyle \frac{\pi}{2}$
Consider \begin{array}{ccccc}
K_n & : & \mathbb R^+ & \to & R\\
 & & t & \mapsto & \arctan(t/x_n)e^{-t} \\
\end{array}
$K_n$ is a continous function for each $n$, and the sequence $(K_n)$ converges pointwise to $\displaystyle t\to \frac{\pi}{2}e^{-t} $, which is continuous as well.
Moreover, $|K_n(x)|\leq \frac{\pi}{2}e^{-t} $ and $\displaystyle t\to \frac{\pi}{2}e^{-t} $ is integrable over $\mathbb R^+$ which yields domination.
Therefore, $\displaystyle \lim_{n\to \infty} \int_{0}^{\infty} K_n(t) dt = \int_{0}^{\infty} \lim_{n\to \infty} K_n(t) dt$
Which can be rewritten as $\displaystyle \lim_{n\to \infty} \int_{0}^{\infty} \arctan(t/x_n)e^{-t} dt = \int_{0}^{\infty} \frac{\pi}{2}e^{-t}$
This, in turn, is equivalent to $\displaystyle \lim_{n\to \infty} f(x_n)= \frac{\pi}{2} \int_{0}^{\infty} e^{-t} = \frac{\pi}{2}$
I've proven that for any sequence of positive numbers $(x_n)$ that goes to $0$, $f(x_n)$ goes to $\frac{\pi}{2}$. This implies that $\displaystyle \lim_{x \rightarrow 0}f(x)=\frac{\pi}{2}$


*

*This is more simple.


It is easy to prove that $\forall y \geq 0, \arctan(y) \leq y$
Hence $\displaystyle \int_{0}^{\infty} \arctan(t/x)e^{-t} dt \leq \int_{0}^{\infty} \frac{t}{x}e^{-t} dt $
So $\displaystyle f(x) \leq \frac{1}{x} \int_{0}^{\infty}te^{-t} dt$
And $\displaystyle 0\leq f(x) \leq \frac{1}{x}$
Taking limits as $x\to \infty$, $\displaystyle \lim_{x \rightarrow \infty}f(x)=0$


*

*$f$ is a decreasing continuous function over $(0,\infty)$.


Hence $\displaystyle \qquad \sup_{x>0} f(x) = \lim_{x \rightarrow 0}f(x)=\frac{\pi}{2}$
And $\displaystyle \qquad \inf_{x>0} f(x) = \lim_{x \rightarrow \infty}f(x)=0$

I leave the last part to you. You just need to justify differentiation under the integral sign.
A: We can rewrite $f(x)$ according to ($u=t/x$)
\begin{eqnarray*}
f(x) &=&\int_{0}^{\infty }dt\exp [-t]\arctan \frac{t}{x}=x\int_{0}^{\infty
}du\exp [-xu]\arctan u \\
&=&-\int_{0}^{\infty }du\{\partial _{u}\exp [-xu]\}\arctan u \\
&=&-\left. \exp [-xu]\arctan u\right\vert _{0}^{\infty }+\int_{0}^{\infty
}du\exp [-xu]\partial _{u}\arctan u \\
&=&\int_{0}^{\infty }du\exp [-xu]\partial _{u}\arctan u=\int_{0}^{\infty
}du\exp [-xu]\frac{1}{1+u^{2}}
\end{eqnarray*}
Now matters become simple.
