Solving an integral that depends on a parameter Can you please help me in solving the following exercise?
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!
 A: 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.
A: 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.
A: 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.
