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if $x > 0$, show that

$$f(x)=\int_0^\infty \frac{\cos(xt)}{1+t^2} \, dt= \frac{\pi}{2}e^{-x}$$

The problem appears in the book Differential Equations: With Applications and Historical Notes, under the chapter derivative and integral of Laplace transformation. So I tried to apply the equation

$$\int_0^\infty \frac {f(t)} t \, dt = \int_0^\infty F(p)\,dp,$$

where $F(p) = L[f(x)](p)$, the Laplace transformation of $f(t)$. However, I failed to manipulate the expression to apply this equality.

I need your help. Please not be misled by my approach, it might be totally wrong.

Elmo
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    use the Fourier inversion theorem : $\displaystyle\int_{-\infty}^\infty e^{-|x|} e^{i \omega x} dx = \frac{1}{1-i\omega}+\frac{1}{1+i\omega} = \frac{2}{1+\omega^2}$ – reuns Oct 17 '16 at 05:05
  • This is definitely related to fourier transform. – polfosol Oct 17 '16 at 05:08
  • The title of your book is "differential equations..." isn't a hint? ;) – FDP Oct 17 '16 at 10:01

1 Answers1

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You can find a lot of ways to prove your equality here. Another way (which I love) is the following. We have that $$I=\int_{0}^{\infty}\frac{\cos\left(xt\right)}{1+t^{2}}dt=\frac{1}{2}\int_{0}^{\infty}\frac{e^{ixt}+e^{-ixt}}{1+t^{2}}dt $$ $$=\frac{1}{4}\int_{0}^{\infty}\frac{e^{ixt}+e^{-ixt}}{1-it}dt+\frac{1}{4}\int_{0}^{\infty}\frac{e^{ixt}+e^{-ixt}}{1+it}dt $$ $$=\frac{e^{-x}}{4i}\int_{0}^{\infty}\frac{ixt\left(e^{x+ixt}+e^{x-ixt}\right)}{t\left(x-ixt\right)}dt+\frac{e^{-x}}{4i}\int_{0}^{\infty}\frac{ixt\left(e^{x+ixt}+e^{x-ixt}\right)}{t\left(x+ixt\right)}dt$$ $$=\frac{e^{-x}}{4i}\int_{0}^{\infty}\frac{ixte^{x+ixt}}{t\left(x+ixt\right)}+\frac{ixte^{x-ixt}}{t\left(x-ixt\right)}dt+\frac{e^{-x}}{4i}\int_{0}^{\infty}\frac{ixte^{x-ixt}}{t\left(x+ixt\right)}+\frac{ixte^{x+ixt}}{t\left(x-ixt\right)}dt $$ so if we apply the complex version of the Frullani's theorem to the functions $$f\left(t\right)=\frac{te^{x-t}}{x-t} $$ and $$g\left(t\right)=\frac{te^{x-t}}{x+t} $$ we get $$I=\frac{e^{-x}}{2i}\log(-1)=\color{red}{\frac{\pi}{2}e^{-x}} $$ as wanted.

Community
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Marco Cantarini
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    (+1) You should change your nickname in Marco Frullani ;) – Jack D'Aurizio Oct 17 '16 at 19:49
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    @JackD'Aurizio It is a good idea but I think I will follow my childhood dream. I will be Max Power (Simpsons docet). – Marco Cantarini Oct 17 '16 at 21:12
  • @ Marco and @JackD'Aurizio well the complex Frullani theorem isn't easier than the Fourier inversion theorem, or the residue theorem. On my side, I wonder if you can find an elementary way to solve $f(s)+f''(s)= \frac{1}{s}$ (coming from $ f(s) = \int_0^\infty \frac{e^{-st}}{1+t^2}dt$) – reuns Oct 18 '16 at 18:56