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 $[a,b]$.
Best Answer
$\newcommand{\d}{\mathop{}\!\mathrm{d}}$
$$\begin{aligned} \int_0^\infty\frac{\sin t}{t+x} \d t &=\int_0^\infty\left(\left.\frac{-e^{-s(t+x)}}{t+x}\right|_{s=0}^\infty\right) \sin t \d t\\ &=\int_0^\infty\left(\int_0^\infty e^{-s(t+x)}\d s\right) \sin t \d t\\ &=\int_0^\infty e^{-sx}\left(\int_0^\infty e^{-st}\sin t\d t\right)\d s\\ &=\int_0^\infty\frac{e^{-sx}}{1+s^2} \d s\\ &=\int_0^\infty\frac{e^{-tx}}{1+t^2} \d t \end{aligned}$$
The second last equality $\int_0^\infty e^{-ts}\sin t \d t=\frac1{1+s^2}$ comes from the following computation. $$\begin{aligned}\int_0^\infty e^{-ts}\sin t \d t &=\int_0^\infty e^{-ts} \d(-\cos t)\\ &=-\left.(\cos t) e^{-ts}\right|_{t=0}^\infty - \int_0^\infty(-\cos t)\d(e^{-ts})\\ &=1 - s\int_0^\infty e^{-ts}\cos t \d t\\ &=1 - s\int_0^\infty e^{-ts}\d(\sin t)\\ &=1 - s\left(\left.e^{-ts}\sin t\right|_{t=0}^\infty - \int_0^\infty \sin t\d (e^{-ts})\right)\\ &=1 - s\left(s\int_0^\infty e^{-ts}\sin t \d t\right) \end{aligned}$$