Continuity of the function $f(y)=\int_{0}^{+\infty} y\sin(x) e^{-xy} \, dx $ in $y=0$

Question

How do you prove the continuity of the follow function in $y=0$

$f:[0,+\infty) \rightarrow \mathbb{R}: f(y) =\int_0^{+\infty} y\sin(x) e^{-xy} \, dx$

We change the variables: $x\rightarrow \frac {x}{y}$ So now we have: $$f(y)=\int_0^{+\infty} \sin\left(\frac{x}{y}\right) e^{-x} \, dx $$

To prove the continuity, we take a row $y_n \rightarrow 0$. We can assume that $y_n>0$ for each $n$. Now we have to prove that $f(y_n) \rightarrow 0$

Now we can apply the dominated convergence theorem with dominated function $e^{-x}$ but then I'm stuck. (the change of variables was given as a hint). Can someone help me.

Answer 1

Hint: $$ f(y)=\int_0^{+\infty} \sin\left(\frac{x}{y}\right) e^{-x} \, dx =\int_0^{+\infty} \sin {(tx)} e^{-x} \, dx\to 0 $$ as $t\to \infty$ ($y\to 0$) by Riemann-Lebesgue lemma, which is proved in How to prove $\lim\limits_{x\to\infty}\int_a^bf(t)\sin(xt)\,dt=0$.

Answer 2

Suppose that you consider $$I=\int y\,\cos(x) e^{-xy} \, dx$$ $$J=\int y\,\sin(x) e^{-xy} \, dx$$ then $$I+iJ=\int y \,e^{-x (y-i)} \, dx=-\frac{y\, e^{-x (y-i)}}{y-i}$$ which makes $$I=\frac{y \,e^{-x y} (\sin (x)-y \cos (x))}{y^2+1}$$ $$J=-\frac{y \,e^{-x y} (y \sin (x)+\cos (x))}{y^2+1}$$ Then $$K=\int_0^\infty y\sin(x) e^{-xy} \, dx=\frac{y}{y^2+1}\qquad \Re(y)>0$$ $$L=\int_0^\infty y\cos(x) e^{-xy} \, dx=\frac{y^2}{y^2+1}\qquad \Re(y)>0$$