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Prove that for any $x > 0, \int_0^\infty \dfrac{e^{-tx}}{1+t^2}dt = \int_0^\infty \dfrac{\sin t}{t+x}dt$.

Let $f(x)$ denote the LHS and $g(x)$ denote the RHS. First of all the LHS is clearly convergent as an integral because for $x>0$ it is at most the value of the integral $\int_0^\infty e^{-tx}dt = 1/x.$ Note that the equality also holds when $x=0$ (though it is fairly nontrivial to show directly that $\int_0^\infty \dfrac{\sin t}t dt = \dfrac{\pi}2$). I know that by Leibniz's integral differentiation rule, we have $f'(x) = \int_0^\infty \dfrac{-t e^{-tx}}{1+t^2}dt.$ I also know that a trick for evaluating integrals is to introduce new variables (e.g. one can introduce a new variable to a double integral and apply Fubini's theorem, which usually helps simplify the integral). One can also use substitution, generalization (e.g. the integral $\int_0^{\pi/2} \dfrac{1}{1+\tan^{\sqrt{2}}(x)} dx$ can be evaluated by observing that $\sqrt{2}$ can be replaced by any real number a), Taylor expansion, partial fraction decomposition, integration by parts, and the Dominated Convergence Theorem. We also have by Leibniz's rule (assuming $g(x)$ actually exists) that $g'(x) = \int_0^\infty -\dfrac{\sin t}{(t+x)^2}dt$. I also know the Fundamental theorem of calculus, which says that $H'(x) = h(x)$ for any (Riemann) integrable function $h:[a,b]\to\mathbb{R}, a<b\in\mathbb{R}$ where $H(x) = \int_a^x h(t)dt$ for all real numbers $x$ in $[a,b]$.

Ѕᴀᴀᴅ
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user3379
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  • $$I(x)=\int_0^\infty\frac{e^{-tx}}{1+t^2}dt=\int_0^\infty\frac{e^{-s}}{x-is}\frac{ds}{2}+\int_0^\infty\frac{e^{-s}}{x+is}\frac{ds}{2}=I_1+I_1;,x>0$$ Both integral converges. We consider quarter-circle contours: in the upper half of complex plane ($I_1$) and low half-plane ($I_2$). There are not poles inside the contours, and the proof that the integrals along the arches tend to zero at $R\to\infty$ is straightforward. Formally it means the substitutions $s=it$ in the first and $s=-it$ in the second integrals correspondingly. – Svyatoslav Oct 09 '22 at 12:45
  • $$I(x)=\int_0^\infty\frac{e^{-it}}{x+t}\frac{idt}{2}+\int_0^\infty\frac{e^{it}}{x+t}\frac{(-i)dt}{2}=\int_0^\infty\frac{\sin t}{x+t}dt$$ – Svyatoslav Oct 09 '22 at 12:45
  • One source: Ukrainian National Mathematical Olympiad for University Students, 1998, problem A3 – Apass.Jack Oct 09 '22 at 18:34

2 Answers2

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$\newcommand{\d}{\mathop{}\!\mathrm{d}}$

$$\begin{aligned} \int_0^\infty\frac{\sin t}{t+x} \d t &=\int_0^\infty\left(\left.\frac{-e^{-s(t+x)}}{t+x}\right|_{s=0}^\infty\right) \sin t \d t\\ &=\int_0^\infty\left(\int_0^\infty e^{-s(t+x)}\d s\right) \sin t \d t\\ &=\int_0^\infty e^{-sx}\left(\int_0^\infty e^{-st}\sin t\d t\right)\d s\\ &=\int_0^\infty\frac{e^{-sx}}{1+s^2} \d s\\ &=\int_0^\infty\frac{e^{-tx}}{1+t^2} \d t \end{aligned}$$

The second last equality $\int_0^\infty e^{-ts}\sin t \d t=\frac1{1+s^2}$ comes from the following computation. $$\begin{aligned}\int_0^\infty e^{-ts}\sin t \d t &=\int_0^\infty e^{-ts} \d(-\cos t)\\ &=-\left.(\cos t) e^{-ts}\right|_{t=0}^\infty - \int_0^\infty(-\cos t)\d(e^{-ts})\\ &=1 - s\int_0^\infty e^{-ts}\cos t \d t\\ &=1 - s\int_0^\infty e^{-ts}\d(\sin t)\\ &=1 - s\left(\left.e^{-ts}\sin t\right|_{t=0}^\infty - \int_0^\infty \sin t\d (e^{-ts})\right)\\ &=1 - s\left(s\int_0^\infty e^{-ts}\sin t \d t\right) \end{aligned}$$

Apass.Jack
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Using your notation, we denote $f(x) = \int_{0}^{\infty} \frac{e^{-tx}}{1+t^2}\, \mathrm{d}t$. Using Leibniz's integral rule we see that $\require{\cancel}$ \begin{align} f '' + f = \int_{0}^{\infty} \frac{\partial^2}{\partial x^2}\frac{e^{-tx}}{1+t^2}\, \mathrm{d}t + \int_{0}^{\infty} \frac{e^{-tx}}{1+t^2}\, \mathrm{d}t = \int_{0}^{\infty} \frac{\cancel{\left(1+t^2 \right)}e^{-tx}}{\cancel{1+t^2}}\, \mathrm{d}t = \frac{1}{x} \end{align} On the other hand, if $g(x) = \int_{0}^{\infty} \frac{\sin(t)}{t+x}\, \mathrm{d}t $ then \begin{align} g'' = \int_{0}^{\infty} \frac{\partial^2}{\partial x^2} \frac{\sin(t)}{t+x}\, \mathrm{d}t = \int_{0}^{\infty} \frac{2\sin(t)}{(t+x)^3}\, \mathrm{d}t \overset{\text{IBP}}{=} -\frac{\sin(t)}{(t+x)^2} - \frac{\cos(t)}{(t+x)}\Bigg\vert_{0}^{\infty} - \int_{0}^{\infty} \frac{\sin(t)}{t+x}\, \mathrm{d}t = \frac{1}{x} - g \end{align} Thus, both $f$ and $g$ satisfy the same differential equation $y'' + y = \frac{1}{x}$. Lastly, since $\lim\limits_{x\to 0^+}f(x) = $$\, \frac{\pi}{2} = \lim\limits_{x\to 0^+}g(x)$ and that $\lim\limits_{x\to \infty}f'(x) = \lim\limits_{x\to \infty}g'(x) =0$ then they also have the same initial conditions, and by the uniqueness of a solution to an ODE we can conclude that $f(x) = g(x)$ for all $x> 0$.

We can actually evaluate both integrals by solving the ODE $y''(x) + y(x) = \frac{1}{x}$. The homogenous equation is $y'' = -y$ which is known to have a sinusoidal solutions $u_1(x) = \sin(x)$ and $u_2(x) = \cos(x)$. Thus, we get the homogeneous solution $$ y_h = c_1\sin(x) + c_2\cos(x) $$ Since the Wronskian is given by $W={\scriptstyle\begin{vmatrix} \sin(x) & \cos(x) \\ \cos(x) & -\sin(x)\end{vmatrix}}=-1$, using the method of variation of parameters we get the particular solution $$ y_p =-u_1\int \frac{u_2 \frac{1}{x}}{W}\mathrm{d}x +u_2\int \frac{u_1\frac{1}{x}}{W}\mathrm{d}x= \sin(x)\int\frac{\cos(x)}{x}\mathrm{d}x - \cos(x)\int\frac{\sin(x)}{x}\mathrm{d}x $$ where we immediately recognize the last two integrals as $\mathrm{Ci}(x)$ and $\mathrm{Si}(x)$. Hence, the general solution is $y =y_h + y_p$. But recalling the initial conditions we get $\frac{\pi}{2} = \lim\limits_{x\to 0^+}y(x) = c_2$ and $0 =\lim\limits_{x\to \infty}y'(x) = c_1$. We thus conclude $$ \boxed{\int_{0}^{\infty} \frac{e^{-tx}}{1+t^2}\, \mathrm{d}t = \int_{0}^{\infty} \frac{\sin(t)}{t+x}\, \mathrm{d}t = \frac{\pi}{2}\cos(x) +\sin(x)\mathrm{Ci}(x) -\cos(x) \mathrm{Si}(x), \ \ x>0} $$

Robert Lee
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