# Dominated Convergence Theorem - variable in set over which we integrate

The Dominated Convergence Theorem (DCT) is extremely useful for computing limits of integrals if the limit variable is in the integrand, so for instance, when computing integrals of the form $$\lim_{n \to \infty} \int f_n~d\lambda$$ But in the following case, the limit variable is in the set over which we integrate: I want to prove that $$\lim_{R \to \infty} \int_{[0, R]} \int_{[0, \infty)} \sin x \cdot e^{-x t} ~ d\lambda(t) ~ d\lambda(x) = \int_{[0, \infty)} \int_{[0, \infty)} \sin x \cdot e^{-x t} ~ d\lambda(t) ~ d\lambda(x)$$ where $\lambda$ is the standard Lebesgue measure.

I thought about doing some trick with indicator functions: we can write $$\lim_{R \to \infty} \int_{[0, R]} \int_{[0, \infty)} \sin x \cdot e^{-x t} ~ d\lambda(t) ~ d\lambda(x) = \lim_{R \to \infty} \int_{[0, R]} 1_{[0, R]} \int_{[0, \infty)} \sin x \cdot e^{-x t} ~ d\lambda(t) ~ d\lambda(x) =$$ $$\lim_{R \to \infty} \int_{[0, \infty)} 1_{[0, R]} \int_{[0, \infty)} \sin x \cdot e^{-x t} ~ d\lambda(t) ~ d\lambda(x) = \lim_{R \to \infty} \int_{[0, \infty)} 1_{[0, R]} \cdot \sin x \int_{[0, \infty)} e^{-x t} ~ d\lambda(t) ~ d\lambda(x) =$$ $$\lim_{R \to \infty} \int_{[0, \infty)} 1_{[0, R]} \cdot \frac{\sin x}{x} ~ d\lambda(x)$$ So all that is left to do is justify switching the limit and the outer integral. But we cannot apply the DCT, since the ''smallest'' function that dominates the absolute values of all of the possible integrand functions (i.e. all integrands for all values of $R$) is $$\left| \frac{\sin x}{x} \right|$$ but we know that that function is not Lebesgue integrable...

How should I prove the desired result?

You can compute the inner integral to get the expression as $$\lim_{R\to\infty}\int_0^R \dfrac{\sin x}{x}dx$$

Now for each $R>1$: $$\int_0^R \dfrac{\sin x}{x} dx=\int_0^1\dfrac{\sin x}{x}dx+\int_1^R\dfrac{\sin x}{x}dx=I_1+I_2$$

Note that since $\lim_{x\to0}\dfrac{\sin x}{x}=1$ it follows that $I_1<\infty$.

Therefore $$I_2=\int_1^R \dfrac{\sin x}{x}dx=\cos 1-\dfrac{\cos R}{R}-\int_1^R\dfrac{\cos x}{x^2}dx$$

Now, note $I_1$ is independent of $R$, so $\lim_{R\to\infty} I_1=I_1$. Further note $\lim_{R\to\infty} \dfrac{\cos R}{R}=0$ and $$\lim_{R\to\infty} \int_1^R\dfrac{\cos x}{x^2}dx=\int_1^\infty\dfrac{\cos x}{x^2}dx\in \left[-\int_1^\infty \dfrac{1}{x^2}dx,\int_1^\infty \dfrac{1}{x^2}dx\right]$$ and we know that $\int_1^\infty \dfrac{1}{x^2}dx<\infty$.

Therefore, $\lim_{R\to\infty} \int_1^R \dfrac{\cos x}{x^2}dx=E<\infty$

So $$\lim_{R\to\infty}\int_0^R \dfrac{\sin x}{x}dx=\int_0^1\dfrac{\sin x}{x}dx+\cos1-E$$ exists and is finite. Plug in these in a computer to get numerical values.