Evaluating $\int_0^\infty\frac{\sin(x)}{x^2+1}\, dx$ I have seen $$\int_0^\infty \frac{\cos(x)}{x^2+1} \, dx=\frac{\pi}{2e}$$ evaluated in various ways. 
It's rather popular when studying CA. 
But, what about $$\int_0^\infty \frac{\sin(x)}{x^2+1} \, dx\,\,?$$
This appears to be trickier and more challenging.
I found that it has a closed form of 
$$\cosh(1)\operatorname{Shi}(1)-\sinh(1)\text{Chi(1)}\,\,,\,\operatorname{Shi}(1)=\int_0^1 \frac{\sinh(x)}{x}dx\,\,,\,\, \text{Chi(1)}=\gamma+\int_0^1 \frac{\cosh(x)-1}{x} \, dx$$
which are the hyperbolic sine and cosine integrals, respectively.
It's an odd function, so 
$$\int_{-\infty}^\infty \frac{\sin(x)}{x^2+1} \, dx=0$$
But, does anyone know how the former case can be done? Thanks a bunch. 
 A: Partial fractions to the rescue!
$$ \frac{1}{x^2 + 1} = \frac{i}{2} \left( \frac{1}{x+i} - \frac{1}{x-i} \right) $$
Then, the angle addition formulas to match the arguments to the denominators
$$ \sin(x) = \sin(x+i) \cos(i) - \sin(i) \cos(x+i) $$
$$ \sin(x) = \sin(x-i) \cos(i) + \sin(i) \cos(x-i) $$
And we can compute
$$ \int_0^\infty \frac{\sin(x+i)}{x+i} \, dx = \int_i^\infty \frac{\sin(x)}{x} \, dx = \frac{\pi}{2} -\text{Si}(i) $$
and similar. Therefore,
$$ \int_0^\infty \frac{\sin x}{x+i} \, dx = \left(\frac{\pi}{2}-\text{Si}(i)\right) \cos(i) +
\sin(i) \text{Ci}(i) $$
$$ \int_0^\infty \frac{\sin x}{x-i} \, dx = \left(\frac{\pi}{2}-\text{Si}(-i)\right) \cos(i) - 
\sin(i) \text{Ci}(-i) $$
Therefore,
$$ \int_0^\infty \frac{\sin(x)}{x^2 + 1}
= \frac{i}{2} \left( \left(-\text{Si}(i) + \text{Si}(-i) \right) \cos(i) + (\text{Ci}(i) + \text{Ci}(-i)) \sin(i) \right) $$
A: Mellin transform of sine is, for $-1<\Re(s)<1$:
$$
   G_1(s) = \mathcal{M}_s(\sin(x)) = \int_0^\infty x^{s-1}\sin(x) \mathrm{d} x =\Im \int_0^\infty x^{s-1}\mathrm{e}^{i x} \mathrm{d} x = \Im \left( i^s\int_0^\infty x^{s-1}\mathrm{e}^{-x} \mathrm{d} x \right)= \Gamma(s) \sin\left(\frac{\pi s}{2}\right) =  2^{s-1} \frac{\Gamma\left(\frac{s+1}{2}\right)}{\Gamma\left(1-\frac{s}{2}\right)} \sqrt{\pi}
$$
And Mellin transfom of $(1+x^2)^{-1}$ is, for $0<\Re(s)<2$:
$$
  G_2(s) =  \mathcal{M}_s\left(\frac{1}{1+x^2}\right) = \int_0^\infty \frac{x^{s-1}}{1+x^2}\mathrm{d} x \stackrel{x^2=u/(1-u)}{=} \frac{1}{2} \int_0^1 u^{s/2-1} (1-u)^{-s/2} \mathrm{d}u = \frac{1}{2} \operatorname{B}\left(\frac{s}{2},1-\frac{s}{2}\right) = \frac{1}{2} \Gamma\left(\frac{s}{2}\right) \Gamma\left(1-\frac{s}{2}\right) = \frac{\pi}{2} \frac{1}{\sin\left(\pi s/2\right)}
$$
Now to the original integral, for $0<\gamma<1$:
$$
    \int_0^\infty \frac{\sin(x)}{1+x^2}\mathrm{d}x = \int_{\gamma-i \infty}^{\gamma+ i\infty} \mathrm{d} s\int_0^\infty \sin(x) \left( \frac{G_2(s)}{2 \pi i} x^{-s}\right) \mathrm{d}s = \frac{1}{2 \pi i} \int_{\gamma-i \infty}^{\gamma+i \infty}  G_2(s) G_1(1-s) \mathrm{d}s =\\ \frac{1}{4 i} \int_{\gamma-i \infty}^{\gamma+i \infty} \Gamma(1-s) \cot\left(\frac{\pi s}{2}\right) \mathrm{d} s = \frac{2\pi i}{4 i} \sum_{n=1}^\infty \operatorname{Res}_{s=2n} \Gamma(1-s) \cot\left(\frac{\pi s}{2}\right) = \sum_{n=1}^\infty \frac{\psi(2n)}{\Gamma(2n)} = \sum_{n=1}^\infty \frac{1+(-1)^n}{2} \frac{\psi(n)}{\Gamma(n)}
$$
Since 
$$
   \sum_{n=1}^\infty z^n \frac{\psi(n)}{\Gamma(n)} = \mathrm{e}^z z \left(\Gamma(0,z) + \log(z)\right)
$$
Combining:
$$
  \int_0^\infty \frac{\sin(x)}{1+x^2} \mathrm{d}x = \frac{\mathrm{e}}{2} \Gamma(0,1) - \frac{1}{2 \mathrm{e}} \Gamma(0,-1) - \frac{i \pi }{2 \mathrm{e}} = \frac{1}{2e} \operatorname{Ei}(1) - \frac{\mathrm{e}}{2} \operatorname{Ei}(-1)
$$
A: Here is another solution:
Consider the integral
$$I(\alpha) = \int_{0}^{\infty} \frac{\sin (\alpha x)}{1+x^2} \, dx = \int_{0}^{\infty} \frac{\alpha \sin x}{\alpha^2+x^2} \, dx.$$
Differentiating $I(\alpha)$ with the first equality, we have
\begin{align*}
I'(\alpha)
&= \int_{0}^{\infty} \frac{x \cos (\alpha x)}{1+x^2} \, dx
 = \int_{0}^{\infty} \frac{x \cos x}{\alpha^2+x^2} \, dx.
\end{align*}
Differentiating once again, we have
\begin{align*}
I''(\alpha)
&= -\int_{0}^{\infty} \frac{2\alpha x \cos x}{(\alpha^2+x^2)^2} \, dx
 = \left[ \frac{\alpha \cos x}{\alpha^2+x^2} \right]_{0}^{\infty} + \int_{0}^{\infty} \frac{\alpha \sin x}{\alpha^2+x^2} \, dx \\
&= -\frac{1}{\alpha} + I(\alpha).
\end{align*}
Thus $I$ satisfies the differential equation
$$ I'' - I = -\frac{1}{\alpha}. \tag{1}$$
To solve this equation, we let
$$ I(\alpha) = u e^{\alpha}. $$
Plugging this to $(1)$ and multiplying $e^{\alpha}$ to both sides, we obtain
$$ (u'e^{2\alpha})' = -\frac{1}{\alpha}e^{\alpha}. $$
Thus integrating both sides, we have
$$ u'e^{2\alpha} = -\mathrm{Ei}(\alpha) - \frac{c_{1}}{2}, $$
where
$$\mathrm{Ei}(\alpha) = PV \int_{-\infty}^{\alpha} \frac{e^{t}}{t} \, dt$$
is the exponential integral function. Then
$$ u' = -e^{-2\alpha}\mathrm{Ei}(\alpha) - \frac{c_{1}}{2}e^{-2\alpha} $$
and hence
\begin{align*}
u
&= \int \left( -e^{-2\alpha}\mathrm{Ei}(\alpha) - \frac{c_{1}}{2}e^{-2\alpha} \right) \, d\alpha \\
&= \frac{1}{2}e^{-2\alpha} \mathrm{Ei}(\alpha) - \int \frac{e^{-\alpha}}{2\alpha} \, d\alpha + c_{1}e^{-2\alpha} + c_{2} \\
&= \frac{1}{2}e^{-2\alpha} \mathrm{Ei}(\alpha) - \frac{1}{2}\mathrm{Ei}(-\alpha) + c_{1}e^{-2\alpha} + c_{2}.
\end{align*}
Therefore it follows that
$$ I(\alpha) = \frac{e^{-\alpha} \mathrm{Ei}(\alpha) - e^{\alpha}\mathrm{Ei}(-\alpha)}{2} + c_{1}e^{-\alpha} + c_{2} e^{\alpha} $$
for some $c_1$ and $c_2$. To determine $c_1$ and $c_2$, observe that
$$\mathrm{Ei}(\alpha) \sim c + \log |\alpha|$$
near $\alpha = 0$. (In fact, we have $c = \gamma$.) Thus taking $\alpha \to 0$,
$$ 0 = I(0) = c_1 + c_2. $$
This shows that we may write
$$ I(\alpha) = \frac{e^{-\alpha} \mathrm{Ei}(\alpha) - e^{\alpha}\mathrm{Ei}(-\alpha)}{2} + c \sinh \alpha. $$
But L'hospital's rule shows that
$$ \mathrm{Ei}(\alpha) \sim \frac{e^{\alpha}}{\alpha} $$
as $|\alpha| \to \infty$. Thus $ I(\alpha) \sim c \sinh \alpha$ as $\alpha \to \infty$. But it is clear that $I(\alpha)$ is bounded:
$$ \left|I(\alpha)\right| \leq \int_{0}^{\infty} \frac{1}{1+x^2} \, dx = \frac{\pi}{2}. $$
Therefore $c = 0$ and we have
$$ \int_{0}^{\infty} \frac{\sin (\alpha x)}{1+x^2} \, dx = \frac{e^{-\alpha} \mathrm{Ei}(\alpha) - e^{\alpha}\mathrm{Ei}(-\alpha)}{2}. $$
A: I don't see how this integral can be evaluated using complex analysis. At some point, you're going to need a circular path with $r \rightarrow \infty$ to go to zero, and the numerator has:
$$
\sin \left(r e^{i\theta}\right) = \frac{1}{2i}\left[\exp\left(i r \cos \theta\right) \exp\left(- r \sin \theta\right) - \exp\left(-i r \cos \theta\right)\exp\left(r \sin \theta\right)\right].
$$
You might look at that and think you can break the integral up into two pieces: the first closed above the $x$ axis so that $\sin \theta > 0$ and the second closed below so that $\sin \theta < 0$. But as you noted, you have to integrate along the positive real axis only (the entire real axis will yield 0), which means you have to use a circular path at $r \rightarrow \infty$ with $\theta$ from $0$ to $2 \pi$.
