# Prove $\int_0^b \left(\int_{0}^\infty f \,dy\right) dx= \int_0^\infty \left(\int_{0}^b f \,dx\right) dy$

I have to prove that for $f(x,y)=e^{-xy^2}\sin(x)$ and $\forall b>0$ we have

$$\int_0^b \left(\int_{0}^\infty f \,dy\right) dx= \int_0^\infty \left(\int_{0}^b f \,dx\right) dy$$

I've tried to solve the first integral but I'm only arriving to indeterminations...

Then, I have to prove that

$$\lim_{b\to \infty} \int_{0}^b \dfrac{\sin x}{\sqrt{x}} = \lim_{b\to \infty} \int_{0}^b \dfrac{\cos x}{\sqrt{x}}=\sqrt{\dfrac{\pi}{2}}$$.

I suppose that's realated with the results of the first integrals and I will be able to solve it when I finally prove the first equation.

• The second equation is ill posed, since as $b$ grows, the right integral will oscillate between $\pm\frac{1}{\sqrt2}$ and the left one between $0$ and $\sqrt2$. The limits are therefore not defined. Jan 21 '15 at 12:37
• But for each $b$ is defined isn't it? Jan 21 '15 at 12:44
• Yes, but it's not for every $b$ that it's true, and certainly not the $b$ that makes it equal to $\sqrt{\frac{\pi}{2}}$. As for your first question, that is a basic fact that is true for all integrable functions, and I suggest that you try to use abstract antiderivatives to prove it, rather than trying to solve it for $f$ specifically. Jan 21 '15 at 12:46
• Fubini's theorem comes to mind. Also, why is this tagged (complex-analysis)? Jan 21 '15 at 13:14
• I dont know, someone edited the question Jan 21 '15 at 13:34

First we note I, II and III.
I. For positive $x$, $$\int_{0}^\infty e^{-xy^2}\sin x \,dy=\frac{\sqrt {\pi}}{2}\cdot\frac{\sin x}{\sqrt{x}}.$$ Use $\int_0^{\infty} e^{-y^2} dy$=$\frac{\sqrt\pi}{2}$.
II. Integration by parts leads to$$\int e^{-xy^2}\sin x dx=-\frac{e^{-xy^2}}{1+y^4} (\cos x+y^2\sin x) .$$
III. For all $a, b$ with $0<a<b<\infty$,$$\int_a^b \left(\int_{0}^\infty f \,dy\right) dx= \int_0^\infty \left(\int_{a}^b f \,dx\right) dy .\tag{1}$$ Since $|e^{-xy^2}\sin x|\le e^{-ay^2}$ for all $x (a\le x<\infty)$ and $\int_0^{\infty} e^{-ay^2}\,dy<\infty$ ,we can change the order of integration.

Now we prove that for any $b>0$,$$\int_0^b \left(\int_{0}^\infty f \,dy\right) dx= \int_0^\infty \left(\int_{0}^b f \,dx\right) dy .\tag{2}$$ Proof. We have $$\int_0^b \left(\int_{0}^\infty f \,dy\right) dx=\int_0^a \left(\int_{0}^\infty f \,dy\right) dx+\int_a^b \left(\int_{0}^\infty f \,dy\right) dx .\tag{3}$$The first term of the right hand side$$\int_0^a \left(\int_{0}^\infty f \,dy\right) dx=\int_0^a\frac{\sqrt{\pi}}{2}\cdot\frac{\sin x}{\sqrt{x}} dx\to 0$$as $a\to 0$. Simirarly we have $$\int_0^\infty \left(\int_{0}^b f \,dx\right) dy= \int_0^\infty \left(\int_{0}^a f \,dx\right) dy+ \int_0^\infty \left(\int_a^b f \,dx\right) dy\tag{4}$$ and the first term of the right hand side $$\int_0^\infty \left(\int_{0}^a f \,dx\right) dy =\int_0^\infty \left( \frac{1-e^{-ay^2}\cos a}{1+y^4}- \frac{y^2e^{-ay^2}\sin a}{1+y^4}\right) dy$$tends to $0$ as $a\to 0$. Therefore, tending $a$ to $0$ in (3) and (4), we have (2) due to (1).

Second question:
From (2),we have $$\frac{\sqrt{\pi}}{2}\int_0^b \frac{\sin x}{\sqrt{x}} dx =\int_0^{\infty} \left(\frac{1}{1+y^4}-\frac{e^{-by^2}}{1+y^4}(\cos b+y^2\sin b)\right) dy .$$ Thus we have $$\frac{\sqrt{\pi}}{2}\int_0^\infty \frac{\sin x}{\sqrt{x}} dx =\int_0^{\infty} \frac{1}{1+y^4} dy$$ and $$\int_0^\infty \frac{\sin x}{\sqrt{x}} dx =\sqrt{\dfrac{\pi}{2}}.$$

The same argument for $f(x,y)=e^{-xy^2}\cos x$ leads to $$\int_{0}^\infty \dfrac{\cos x}{\sqrt{x}}=\sqrt{\dfrac{\pi}{2}}.$$

• Thank you!! I did it yesterday and when I arrived to $\int \frac{1}{1+y^4}$ it was a chaos to solve it. Finally I computed it with Wolfram Alpha. Do you think that there is an easy way to did it? Jan 22 '15 at 10:58