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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.

  • 1
    According to the integral book that I have (by Gradshteyn Ryzhik): \begin{equation} \int_0^{\infty}{\frac{x^{\mu-1}\sin(ax)}{1+x^2}} = \frac{\pi}{2}\sec\frac{\mu\pi}{2}\sinh(a) + \\ \frac{1}{2}\sin\frac{\mu\pi}{2}\Gamma(\mu) \left\{\exp\left[-a+i\pi(1-\mu)\right] \gamma(1-\mu, -a) - e^a\gamma(1-\mu,a) \right\} \end{equation} which gives the answer to the question for the case $a=1$ and $\mu=1$. I honestly don't want to spend time to check if it matches to the answers by other people.2015-06-12

4 Answers 4

14

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) $

  • 0
    Nice answer! (+1)2013-01-19
10

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}. $

4

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) $

1

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 $2 \pi$.