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Let $F : [0, + \infty [ \to \mathbb{R}$ be the function defined by $F(x)=\int_0^{+\infty}\frac{e^{-tx}}{1+t^2} \mathrm{dt}$. I'm asked to prove that this function is $\mathcal{C}^2$ on $[0, + \infty[$.

On $]0, + \infty[$, I can prove it by proving that the function is $\mathcal{C}^2$ on each interval $[a, b]$, with $0 < a < b$, and I get $F'(x)=\int_0^{+\infty}\frac{-te^{-tx}}{1+t^2} \mathrm{dt}$, $F''(x)=\int_0^{+\infty}\frac{t^2e^{-tx}}{1+t^2} \mathrm{dt}$.

However, at the point $0$, I'm inclined to believe that the function is not differentiable (and that there is a mistake in the statement...) : for $x=0$, the integral $\int_0^{+\infty}\frac{-te^{-tx}}{1+t^2} \mathrm{dt}$ is not convergent...

Am I right ? If so, how could I prove it rigorously ? If not, how could I prove that the function is differentiable at the point $0$ ?

Thanks in advance !

antwomorfisme
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2 Answers2

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As stated in this answer, we know that $$F(x)=\int_{0}^{+\infty}\frac{e^{-tx}}{1+t^{2}}dt=\mathrm{Ci}\left(x\right)\sin\left(x\right)+\frac{\pi\cos\left(x\right)}{2}-\mathrm{Si}\left(x\right)\cos\left(x\right)\qquad(\blacktriangle)$$ where

$$\mathrm{Ci}(x)=-\int_x^\infty\frac{\cos t}{t}dt$$ and $$\mathrm{Si}(x)=\int_0^x\frac{\sin t}{t}dt$$

are, respectively, the cosine and the sine integrals. The power series expansions for $\mathrm{Ci}(x)$ and $\mathrm{Si}(x)$ are
$$\mathrm{Ci}(x)=\gamma+\ln x+\sum_{k=1}^\infty\frac{(-x^2)^k}{2k(2k)!}$$ and $$\mathrm{Si}(x)=\sum_{k=1}^\infty(-1)^{k-1}\frac{x^{2k-1}}{(2k-1)(2k-1)!}$$ where $\gamma$ is the Euler-Mascheroni constant. They follow directly from their definitions in terms of cosine and sine and the fact that $$\mathrm{Ci}(x)=\gamma+\ln(x)+\int_0^x\frac{\cos t-1}{t}dt$$

At this point it's not difficult to prove that the first terms of the series expansion at $x=0$ of $F(x)$ are $$F(x)=\frac{\pi}{2}+x(\ln x+\gamma-1)-\frac\pi4 x^2+O(x^3)$$ Since $$F(0)=\int_0^{+\infty}\frac{1}{1+t^2}dt=\frac\pi2$$ the incremental ratio of $F$ at $0$ satisfies $$\frac{F(x)-F(0)}{x}=\ln x+\gamma-1-\frac\pi4x+O(x^2)$$ So $$\lim_{x\to 0^+}\frac{F(x)-F(0)}{x}=-\infty$$ which proves that your function $F$ is not differentiable at $0$.

In order to prove $(\blacktriangle)$ it's sufficient to adapt the answer given by Random Variable here. Precisely: \begin{align*} F(x)&=\int_0^\infty\frac{e^{-tx}}{1+t^2}dt \\&=\int_0^\infty\int_0^\infty e^{-tx}\sin(a)e^{-ta}da\:dt \\&=\int_0^\infty\int_0^\infty \sin(a)e^{-(a+x)t}dt\:da \\&=\int_0^\infty\frac{\sin(a)}{a+x}da \\&=\star \end{align*} Changing variable $u=a+x\:\Rightarrow\:du=da$ and remembering the difference formula for the sine $\sin(u-x)=\sin u\cos x-\cos u\sin x$, the integral $\star$ becomes \begin{align} \star&=\cos x\int_x^\infty\frac{\sin u}{u}du-\sin x\int_x^\infty\frac{\cos u}{u}du \\&=\cos x\left(\int_0^\infty\frac{\sin u}{u}du-\int_0^x\frac{\sin u}{u}du\right)+\sin x\left(-\int_x^\infty\frac{\cos u}{u}du\right) \\&=\cos x\left(\frac\pi2-\mathrm{Si}(x)\right)+\sin x\:\mathrm{Ci}(x) \end{align} since $\int_0^\infty\frac{\sin u}{u}du=\frac\pi2$.

user773458
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$$I=\int\frac{e^{-tx}}{1+t^2} \,{dt}=\frac i 2 \int \Big(\frac{e^{-t x}}{t+i}-\frac{e^{-t x}}{t-i}\Big)\,dt$$ Using obvious changes of variables $$I=\frac i 2 \int \Big(\frac{e^{-(u-i) x}}{u}-\frac{e^{-(u+i) x}}{u}\Big)\,du$$ and we face exponential integral functions $$\int \frac{e^{-(u+a) x}}{u}\,du=e^{-a x}\int \frac{e^{-u}}u \,du==e^{-a x}\,\,\text{Ei}(-u x)$$

Recombining $$I=\frac i 2\Big( e^{i x} \text{Ei}(-((t+i) x))-e^{-i x} \text{Ei}(-((t-i) x)) \Big)$$

Using the bounds and the relation betwenn the exponential integral function and the triginometric integrals (assuming $x>0$) $$J(x)=\int_0^\infty\frac{e^{-tx}}{1+t^2} \,{dt}=\text{Ci}(x) \sin (x)-\text{Si}(x) \cos(x)+\frac{\pi}{2} \cos (x)$$ $$J'(x)=\text{Ci}(x) \cos (x)+\text{Si}(x) \sin (x)-\frac{\pi }{2} \sin (x)$$ Expanded around $x=0$ $$J'(x)=\log (x)+\gamma -\frac{\pi x}{2}+O\left(x^2\right)$$

Claude Leibovici
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