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The question says to show (via IBP):
$$\int_{0}^\infty e^{-xy}\sin y \mathrm{d}y = \frac{1}{1+x^2} \quad \text{for} \quad x>0.$$

The solutions did it by first differentiating the exponential and integrating the trigonometric terms... but I did it the other way yet it doesn't give me the answer!?

Here is what I did:
Let $I$ be the integral
$u = \sin y$ and $dv = e^{-xy}\mathrm{d}y$ then
$\mathrm{d}u = \cos y \mathrm{y}$ and $\frac{e^{-xy}}{-x}$

$$I = \left[\sin y\frac{e^{-xy}}{-x}\right]_{0}^\infty - \int_{0}^\infty \frac{e^{-xy}\cos y}{-x} dy\\ \\= \frac{1}{x}\int_{0}^\infty e^{-xy}\cos y \mathrm{d}y$$

Letting
$u = \cos y$ and $dv = e^{-xy}dy$
$du = -\sin y$ and $v = \frac{e^{-xy}}{-x}$

$$I = \frac{1}{x}\left(\left[\cos y \frac{e^{-xy}}{-x}\right]_{0}^\infty - I\right)\\ \\ I = \frac{1}{x}\left(\frac{1}{x} - I\right)\\ \implies I = \frac{1}{x^2} - \frac{1}{x}I \\ \implies \left(1+\frac{1}{x}\right)I = \frac{1}{x^2}\\ \implies I = \frac{1}{x(x+1)}.$$

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    $\begingroup$ The easy way is to compute it as the imaginary part of $\int_0^\infty e^{-(x-i)y}\,dy$. $\endgroup$ Jun 11 '17 at 5:21
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You just missing $\frac1x$ $$I = \left[\sin y\frac{e^{-xy}}{-x}\right]_{0}^\infty - \int_{0}^\infty \frac{e^{-xy}\cos y}{-x} dy\\ \\= \frac{1}{x}\int_{0}^\infty e^{-xy}\cos y \mathrm{d}y\\=\frac1x(\left[\cos y\frac{e^{-xy}}{-x}\right]_{0}^\infty- \int_{0}^\infty \frac{e^{-xy}(-\sin y)}{-x} dy)$$so $$I=\frac1x(\frac1x-\frac1x \int_{0}^\infty \frac{e^{-xy}(+\sin y)}{1} dy) \to I=\frac1x(\frac1x-\frac1xI)\to\\ I=\frac{1}{x^2}(1-I)\\I(1+\frac{1}{x^2})=\frac{1}{x^2}\\I=\frac{1}{1+x^2}$$
Second solution : $$e^{iy}-e^{-iy}=cos y+i\sin y -(cos y-i\sin y)=2i\sin y \to \\\sin y =\dfrac{e^{iy}-e^{-iy}}{2i}$$ $$\int_{0}^\infty e^{-xy}\sin y \mathrm{d}y =\int_{0}^\infty e^{-xy}\dfrac{e^{iy}-e^{-iy}}{2i} \mathrm{d}y =\\ \dfrac{1}{2i}\int_{0}^\infty (e^{y(i-x)}-e^{y(-i-x)})\mathrm{d}y=\ \dfrac{1}{2i} (\dfrac{e^{y(i-x)}}{i-x}-\dfrac{e^{y(-i-x)}}{-i-x})|^{\infty}_0=\\\frac{1}{2i}(0-\frac{1}{i-x}+0-\frac{1}{i+x})=\\ \frac{1}{2i}(-\frac{i+x+i-x}{i^2-x^2}) =\\\frac{1}{2i}(-\frac{2i}{-1-x^2})=\\\frac{1}{1+x^2}$$

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