Solving an integral that depends on a parameter

For any y in $$\mathbb{R}$$ compute the integral: $$F(x) = \int_0^\infty e^{-x} \frac{\sin(xy)}{x}\,dx$$

I tried it by using the partial derivative with respect to y:

$$\frac{\partial F(x)}{\partial y} = \int_0^\infty \frac{\partial f}{\partial y} e^{-x} \frac{\sin(xy)}{x}\,dx$$

$$\frac{\partial F(x)}{\partial y} = \int_0^\infty \frac{\cos(xy)}{e^{x}}\,dx$$

But after that, I don't know how to proceed... Could you please tell me, if my first step was correct or not and how to proceed after this first step? Thank you!

• Did you mean to integrate over $y$, perhaps? I'm not sure it makes sense to have $x$ as the parameter and the integration variable... – emprice Apr 1 at 21:41
• @emprice: If you integrate over $y$, the original integral won't converge. I think it's meant to be $F(y)$ everywhere instead of $F(x)$. – Troposphere Apr 1 at 21:44

The subsequent integral is amenable to integration by parts using a familiar trick. Let $$I(x) = \int e^{-x} \cos yx \, dx.$$ Then with the choice $$u = \cos yx, \quad du = -y \sin yx \, dx, \\ dv = e^{-x} \, dx, \quad v = -e^{-x},$$ we obtain $$I(x) = -e^{-x} \cos yx - y \int e^{-x} \sin yx \, dx.$$ Repeating this process with $$u = \sin yx, \quad du = y \cos yx \, dx, \\ dv = e^{-x} \, dx, \quad v = -e^{-x},$$ we get $$I(x) = -e^{-x} \cos yx + y e^{-x} \sin yx - y^2 \int e^{-x} \cos yx \, dx = e^{-x} (y \sin yx - \cos yx) - y^2 I(x).$$ Therefore, $$I(x) = \frac{e^{-x} (y \sin yx - \cos yx)}{y^2 + 1} + C.$$ The definite integral is then $$\int_{x=0}^\infty e^{-x} \cos yx \, dx = \frac{1}{y^2+1}.$$ I leave the rest as an exercise.
You've made progress: The $$x$$ in the denominator is gone. If you now unfold the cosine in $$e^{-x}\cos(xy)$$ using Euler's formulas, you'll end up with two integrals of the form $$\int \frac12e^{(-1\pm iy)x}\,dx$$, which should be easy to solve.
Here's a more scenic route $$\mathcal L\left\{\frac{\sin(at)}{t}\right\}(s)$$ $$= \int_0^\infty\frac{\sin(at)}{t}e^{-st}\ \mathrm dt$$ $$=\int_s^\infty\int_0^\infty\sin(at)e^{-ut}\ \mathrm dt\ \mathrm du$$ $$=\int_s^\infty\frac{a}{u^2+a^2}\ \mathrm du$$ $$=\int_\frac sa^\infty\frac{1}{w^2+1}\ \mathrm dw$$ $$=\frac\pi 2-\arctan\frac sa$$ $$=\arctan\frac as$$ Now just set $$s=1$$ and we're done.