# Study of a function and other facts

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

1. Prove that $f$ is continuous, monotone and convex.
2. Evaluate: $$\lim_{x \rightarrow 0}f(x), \qquad \lim_{x \rightarrow +\infty}f(x), \qquad \sup_{x>0} f(x), \qquad \inf_{x>0} f(x).$$

3. Show that $f$ is differentiable and find $f'$.

My try:

1. 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?
• what are your ideas on it so far?
– S L
Commented May 24, 2014 at 7:15
• @SantoshLinkha updated with my considerations… Commented May 24, 2014 at 7:48

• 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.

• what the 1 is due to? Commented May 24, 2014 at 9:27
• @TheMaker94 it's the very definition of the convexity of $g_t$ Commented May 24, 2014 at 9:29
• Oh yes, so stupid! Thanks, really clear the first point! Commented May 24, 2014 at 9:33
• @TheMaker94 I wrote the rest of the solution. Commented May 24, 2014 at 9:51
• @Brontolo consider accepting my answer Commented Nov 1, 2017 at 8:04

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.