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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!

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  • $\begingroup$ 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... $\endgroup$ – emprice Apr 1 at 21:41
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    $\begingroup$ @emprice: If you integrate over $y$, the original integral won't converge. I think it's meant to be $F(y)$ everywhere instead of $F(x)$. $\endgroup$ – Troposphere Apr 1 at 21:44
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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.

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  • $\begingroup$ Thanks a lot! I didn't get that I have to integrate by parts multiple times. $\endgroup$ – LStar 16 Apr 3 at 12:02
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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.

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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.

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