# Finding limit of Double Integral

I'm working through some old qual problems and ran into this one that has stumped me. The problem is to compute $$\lim_{n\to \infty} \int_0^\infty \int_0^\infty \frac{n}{x} \sin\left( \frac{x}{ny} \right) e^{-\frac{x}{y} - y}\ dx\ dy.$$

My initial thought was to use Fubini's Theorem to flip the order of integration. Then we get the integral: $$\lim_{n\to \infty} \int_0^\infty\frac{n}{x} \int_0^\infty \sin\left( \frac{x}{ny} \right) e^{-\frac{x}{y} - y}\ dy\ dx.$$

I was hoping to simplify this down to only one integral (in terms of $$x$$) and then apply either MCT or DCT to swap limit and integral, but I am completely lost of where to do with simplifying this integral. I've tried thinking of a helpful change of variable or using integration by parts, but am getting nowhere. Does anyone have any ideas of where to go with this problem?

• Try $sin\frac {x}{ny}=\exp(\frac {ix}{ny})$ as a substitute. Jun 21 at 22:21

DCT is applicable directly, giving the limit as $$\int_0^\infty\int_0^\infty(1/y)e^{-y-x/y}\,dx\,dy$$ (the integrand here is a dominating function for DCT). After integrating over $$x$$, we get $$\int_0^\infty e^{-y}\,dy$$...
Alternatively, substitute $$x=yz$$ in the (inner) integral as written first. You get $$\int_0^\infty\int_0^\infty\frac{n}{z}\sin\frac{z}{n}\,e^{-z-y}\,dz\,dy=\int_0^\infty\frac{n}{z}\sin\frac{z}{n}\,e^{-z}\,dz.$$ Now apply DCT.
• Could you explain why DCT is applicable directly? Also, I think there is an error in the second approach you mentioned. If you substitute $x=yz$ into the integral, you get $$\int_0^\infty \int_0^\infty \frac{n}{yz} \sin\frac{z}{n}e^{-z-y}dzdy.$$ Jun 22 at 15:36
• Edited. When doing $x=yz$ in the inner integral, $y$ serves as a constant. Jun 22 at 15:43
• @halestorm818: (No, you have missed $dx=\color{red}{y}\,dz$.) Jun 22 at 15:45