(I am working on problems having to do with Fubini’s Theorem)
Given $ α ∈ (0,∞)$ , show that the function $ (x, y) \mapsto e^{−αxy}\cdot sin x$ is Lebesgue integrable on $ (0,∞) × (1,∞)$ . Compute the two iterated integrals and use the result to compute
$ \int_0^{\infty} e^{\alpha x} \frac{sinx}{x}dx $
How do I show the function is Lebesgue integrable? Usually I need to show that the Lebesgue integral is finite… but I am new to having two variables in these problems.
Now, for evaluating the integral. I have evaluated each of them below, then set them equal, as the iterated integrals should be equal. Is that correct?
dxdy
$ \int_1^{\infty} \int_0^{\infty} e^{-\alpha xy} \cdot sinx dxdy $
$ I = \int_0^{\infty} e^{-\alpha xy} \cdot sinx dx$
Let $ u = e^{-\alpha yx}, du = -\alpha ye^{-\alpha yx}, v = -cosx, dv = sinxdx$ .
$ I = -cosxe^{-\alpha yx}\rvert_0^{\infty} – \alpha y \int_0^{\infty}cosxe^{-\alpha yx}dx $
Let $ u = e^{-\alpha yx}, du = -\alpha ye^{-\alpha yx}, v = sinx, dv = cosxdx$ .
$ I = (0-(-1)(1)) – \alpha y [e^{-\alpha yx}sinx\rvert_0^{\infty} + \alpha y \int_0^{\infty} e^{-\alpha xy} \cdot sinx dx] $
$ I = 1 – \alpha y(0-0) – \alpha^2 y^2 I$
$ I = \frac{1}{1+\alpha^2 y^2}$
Now we have,
$ \int_1^{\infty} \frac{1}{1+\alpha^2 y^2}$
Let $ u = \alpha x, du = \alpha dx$ .
$ = \frac{1}{\alpha} \int_{\alpha}^{\infty} \frac{1}{1+u^2} du = \frac{1}{\alpha} (arctan(\alpha x))\rvert_1^{\infty} = \frac{1}{\alpha} (\frac{\pi}{2} – arctan(\alpha))$
dydx
$ \int_0^{\infty} \int_1^{\infty} e^{-\alpha xy} \cdot sinx dydx $
$ =\int_0^{\infty} [\frac{sinx}{-\alpha x} \cdot e^{-\alpha xy}]\rvert_1^{\infty} dx = \int_0^{\infty} \frac{sinx}{-\alpha x} (0 – e^{-\alpha x}) dx = \frac{1}{\alpha} \int_0^{\infty} e^{-\alpha x} \cdot \frac{sinx}{x} dx$
Then I set them equal to evaluate the integral the problem asks for.
$ \frac{1}{\alpha} (\frac{\pi}{2} – arctan(\alpha)) = \frac{1}{\alpha} \int_0^{\infty} e^{-\alpha x} \cdot \frac{sinx}{x} dx$
$ \implies \frac{\pi}{2} – arctan(\alpha) = \int_0^{\infty} e^{-\alpha x} \cdot \frac{sinx}{x} dx$
My issue is that in the problem is is $ \alpha x$ not $ -\alpha x$ .
