Compute $\int_0^\infty\frac{\cos(xt)}{1+t^2}dt$

Question

Let $x \in \mathbb R$

Find a closed form of $\int_0^\infty\dfrac{\cos(xt)}{1+t^2}dt$ .

Let me give some context: this is an exercise from an improper integrals course for undergraduates. My teacher was not able to give a valid solution without resorting to Fourier series.

He actually wanted to prove first that the function is a solution of the ODE $$ y-y'' =0$$ without succeeding in proving it.

Anyway the result sould be $\dfrac{\pi}{2}e^{-x}$

Thanks for any help on this subject.

Answer 1

We can also find the differential equation by just differentiating under the integral sign, after rewriting a little:

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

is continuous by the dominated convergence theorem, with $F(0) = \frac{\pi}{2}$ known and $\lvert F(x)\rvert \leqslant \frac{\pi}{2}$. For $x \neq 0$, we can integrate by parts,

$$\begin{align} F(x) &= \left[\frac{1}{x}\cdot \frac{\sin (xt)}{1+t^2}\right]_0^\infty + \frac{2}{x} \underbrace{\int_0^\infty \frac{t\sin (xt)}{(1+t^2)^2}\,dt}_{G(x)}\\ &= \frac{2}{x}G(x). \end{align}$$

$G(x)$ can be differentiated under the integral sign by the dominated convergence theorem, and

$$G'(x) = \int_0^\infty \frac{t^2\cos (xt)}{(1+t^2)^2}\,dt = F(x) - \underbrace{\int_0^\infty \frac{\cos (xt)}{(1+t^2)^2}\,dt}_{H(x)}.$$

$H(x)$ can also be differentiated under the integral sign by the dominated convergence theorem, and

$$H'(x) = -G(x).$$

Thus

$$\begin{align} F'(x) &= -\frac{2}{x^2}G(x) + \frac{2}{x}G'(x)\\ &= -\frac{1}{x}F(x) + \frac{2}{x}\left(F(x) - H(x)\right)\\ &= \frac{1}{x}F(x) - \frac{2}{x}H(x),\quad\text{and}\\ F''(x) &= -\frac{1}{x^2}F(x) + \frac{1}{x} F'(x) + \frac{2}{x^2}H(x) - \frac{2}{x}H'(x)\\ &= -\frac{1}{x^2}F(x) + \frac{1}{x^2}F(x) - \frac{2}{x^2}H(x) + \frac{2}{x^2}H(x) + \frac{2}{x}G(x)\\ &= \frac{2}{x}G(x)\\ &= F(x). \end{align}$$

From the boundedness of $F$ and the differential equation together with the initial value $F(0)$ we then obtain

$$F(x) = \frac{\pi}{2}e^{-\lvert x\rvert}.$$

Answer 2

For the sake of those who might want to see this one solved, I'll sketch the contour integral solution anyway, focusing on $x>0$ for simplicity.

Using a semicircular contour $\Gamma$ in the upper half of the complex plane, centered at the origin, with radius $R$ we get: $$ \oint_\Gamma \frac{e^{ixz}}{1+z^2}\mathrm{d}z = 2\pi i \Re es\left(\frac{e^{ixz}}{1+z^2}\right)_{z=i} $$ but $$ \Re es\left(\frac{e^{ixz}}{1+z^2}\right)_{z=i} = \frac{e^{x i ^2}}{2i} = \frac{e^{-x}}{2i}. $$Considered that as $R\to \infty$ the contribution of the line integral of the upper semicircle vanishes (by Jordan's Lemma), leaving only the one from the real axis integral: $$ \int_{-\infty} ^{\infty} \frac{e^{ixt}}{1+t^2}\mathrm{d}t = 2\pi i \frac{e^{-x}}{2i} = \pi e^{-x}. $$ Now $$ \int_{0} ^{\infty} \frac{\cos{xt}}{1+t^2}\mathrm{d}t =\frac{1}{2}\Re e \int_{-\infty} ^{\infty} \frac{e^{ixt}}{1+t^2}\mathrm{d}t = \frac{\pi}{2}e^{-x}. $$

Answer 3

Let us instead study $$ f(x,a)=\int_0^\infty \frac{\mathrm{e}^{-at}\cos (xt)}{1+t^2}dt. $$ For the above integral, differentiation of the integrand with respect to $x$ is definitely permitted, due to Lebesgue Dominated Convergence Theorem. We have $$ \frac{\partial }{\partial x}f(x,a)=-\int_0^\infty \frac{t\mathrm{e}^{-at}\sin (xt)}{1+t^2}dt \quad\text{and}\quad \frac{\partial^2 }{\partial x^2}f(x,a)=-\int_0^\infty t^2\frac{\mathrm{e}^{-at}\cos (xt)}{1+t^2}dt $$ Therefore $$ f(x,a)-\frac{\partial^2 }{\partial x^2}f(x,a)=\int_0^\infty\mathrm{e}^{-at} \cos(xt)\,dt=\cdots=\frac{a}{a^2+x^2} $$ So, the limit of $f(x,a)$, as $a\to 0$ exists, and does the limit of $\dfrac{a}{a^2+x^2}$, and the latter is equal to zero. Thus$^*$ the limit of $f_{xx}(x,a)$ also exists and, for $a=0$ we have $$ f(x,0)-\frac{\partial^2 }{\partial x^2}f(x,0)=0, $$ Thus $$ \int_0^\infty \frac{\cos(tx)}{1+t^2}dt=f(x,0)=c_1\mathrm{e}^{x}+c_2\mathrm{e}^{-x}, $$ for suitable constants $c_1,c_2$. Clearly $c_1=0$, since $f(x,0)$ remains bounded by $\pi/2$, as $x\to\infty$, and $c_2=\pi/2$, since $$ f(0,0)= \int_0^\infty\frac{dt}{1+t^2}=\frac{\pi}{2}. $$ Thus $$ \int_0^\infty\frac{\cos xt\,dt}{1+t^2}=\frac{\pi\mathrm{e}^{-x}}{2}, $$ which holds only for $x>0$. In particular, the formula which covers all values of $x$ is $$ \int_0^\infty\frac{\cos xt\,dt}{1+t^2}=\frac{\pi\mathrm{e}^{-|x|}}{2}, $$ $^*$ Justification. Using integration by parts we obtain that $$ f(x,0)=\int_0^\infty \frac{\cos xt}{1+t^2}dt=\frac{1}{x}\int_0^\infty\frac{2t\sin tx\,dt}{(1+t^2)^2}=\frac{1}{x^2}\int_0^\infty (\cos tx-1)\left(\frac{2t}{(1+t^2)^2}\right)'\,dt, $$ which shows that $g(x)=f(x,0)$ is twice continuously differentiable, for $x\ge 0$.