Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
How did you find $\cosh(1)\text{Shi(1)}−\sinh(1)\text{Chi}(1)$? – draks ... Sep 5 '12 at 22:24
First look at $\int \frac{e^{ix}}{x^2+1}dx$, ask Wolfram to get $\int \frac{e^{ix}}{x^2+1}dx= \frac{i(e^2\text{Ei(ix-1)}-\text{Ei}(ix+1))}{2e}+const.$. Plug in the limits to get $\int_0^\infty \frac{e^{ix}}{x^2+1}dx=\gamma+i0.64676...$. Then $\int_0^\infty \frac{\sin(x)}{x^2+1}dx=0.64676...$ – draks ... Sep 5 '12 at 22:40
This comes from the antiderivative, $$-\frac{i}{2} \left( {\it Si} \left( x-i \right) \cosh \left( 1 \right) +i{ \it Ci} \left( x-i \right) \sinh \left( 1 \right) \right) +\frac{i}{2} \left( {\it Si} \left( x+i \right) \cosh \left( 1 \right) -i{\it Ci} \left( x+i \right) \sinh \left( 1 \right) \right) $$ which in turn comes from expanding $1/(x^2+1)$ in partial fractions. – Robert Israel Sep 5 '12 at 22:45
Draks. I got the solution from here. Scroll down to 1.5:… I ran it through Maple and pretty much got the solution Robert posted. Thanks everyone for the responses. – Cody Sep 6 '12 at 20:04
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. – Ali Jun 12 '15 at 2:02
up vote 9 down vote accepted

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: $$ \int_0^\infty \frac{\sin(x)}{1+x^2} \mathrm{d}x = \frac{\mathrm{e}}{2} \Gamma(0,1) - \frac{1}{2 \mathrm{e}} \Gamma(0,-1) - \frac{i \pi }{2 \mathrm{e}} = \frac{1}{2e} \operatorname{Ei}(1) - \frac{\mathrm{e}}{2} \operatorname{Ei}(-1) $$

share|cite|improve this answer
Very nice, Sasha. – Cody Sep 7 '12 at 16:16
@Sasha: Melin transform seems to be very useful. I didn't use it so far, but I plan to do it. (+1) – I'm an artist Sep 8 '12 at 6:16
Nice answer! (+1) – Sangchul Lee Jan 19 '13 at 5:06

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

$$ u'e^{2\alpha} = -\mathrm{Ei}(\alpha) - \frac{c_{1}}{2}, $$


$$\mathrm{Ei}(\alpha) = PV \int_{-\infty}^{\alpha} \frac{e^{t}}{t} \, dt$$

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

$$ \int_{0}^{\infty} \frac{\sin (\alpha x)}{1+x^2} \, dx = \frac{e^{-\alpha} \mathrm{Ei}(\alpha) - e^{\alpha}\mathrm{Ei}(-\alpha)}{2}. $$

share|cite|improve this answer

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,

$$ \int_0^\infty \frac{\sin x}{x+i} \, dx = \left(\frac{\pi}{2}-\text{Si}(i)\right) \cos(i) + \sin(i) \text{Ci}(i) $$ $$ \int_0^\infty \frac{\sin x}{x-i} \, dx = \left(\frac{\pi}{2}-\text{Si}(-i)\right) \cos(i) - \sin(i) \text{Ci}(-i) $$


$$ \int_0^\infty \frac{\sin(x)}{x^2 + 1} = \frac{i}{2} \left( \left(-\text{Si}(i) + \text{Si}(-i) \right) \cos(i) + (\text{Ci}(i) + \text{Ci}(-i)) \sin(i) \right) $$

share|cite|improve this answer

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 $2 \pi$.

share|cite|improve this answer
Thank you, Eric. – Cody Sep 7 '12 at 16:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.