How does one calculate a derivative from improper integral like this?
$$\int_t^{\infty}\frac{\sin(x-t)}{x}dx\qquad\ $$
It's been said that this particular integral (as a function) satisfies the equation
$$y''+y=\frac{1}{t},\qquad t>0$$
but I just can't see how.
It's an application of the Leibniz theorem: derivation under the sign $\int$
First by an obvious change of variable we see that this integral is equal to $$y(t)=\int_0^\infty\frac{\sin x}{x+t}dx$$ so let $h(t,x)=\frac{\sin x}{x+t}$ then we have $$\frac{\partial h}{\partial t}(t,x)=-\frac{\sin x}{(x+t)^2}$$ and
$$\frac{\partial^2 h}{\partial t^2}(t,x)=2\frac{\sin x}{(x+t)^3}$$ and we have $$\left|\frac{\partial h}{\partial t}(t,x)\right|\le\frac{1}{(x+a)^2}\in L^1([a,+\infty[),\;\forall a>0$$ and $$\left|\frac{\partial^2 h}{\partial t^2}(t,x)\right|\le\frac{2}{(x+a)^3}\in L^1([a,+\infty[),\;\forall a>0$$ hence $y$ is twice differentiable on $(0,+\infty)$ (in fact $C^\infty(0,+\infty)$) and $$y''(t)=2\int_0^\infty\frac{\sin x}{(x+t)^3}dx$$
Now from the expression of $y$ we integrate by parts and we find $$y=-\frac{\cos x}{t+x}\bigg|_0^\infty-\int_0^\infty \frac{\cos x}{(t+x)^2}dx=\frac1t-\int_0^\infty \frac{\cos x}{(t+x)^2}dx$$ and by a second integration by parts we recognize the expression of $y''$ and we find easily that $$y''+y=\frac1t$$