Computing the integral of the sine integral remainder function Define $\mathrm{si}(x) := \displaystyle\int_x^\infty \frac{\sin(t)}{t}\mathrm{d}t $ for all $x&gt;0$. I have showed, by integration by parts, that this function has convergent integral over $\mathbf{R}^*_+$, but I now have trouble actually computing
$$ \int_0^\infty \mathrm{si}(x)\mathrm d x. $$
Anyone got any pointers ?

Update: For detail, I'll show's $\mathrm{si}$ integral converges. We have :

$$\mathrm{si}(x) = \cos(x)/x - \int_x^\infty \frac{\cos(t)}{t^2}\mathrm d t = \frac{\cos(x)}{x} + \frac{\sin(x)}{x^2} - 2\int_x^\infty\frac{\sin(t)}{t^3}\mathrm d t. $$
The second integral is smaller than the integral of $1/t^3$ over the same interval, which is $1/2x^2$. Thus
$$ \mathrm{si}(x) = \frac{\cos(x)}x + \frac{\sin(x)}{x^2} + O\left(\frac{1}{x^2}\right) = \frac{\cos(x)}x + O\left(\frac{1}{x^2}\right). $$
Integrating by parts, $$ \begin{align} \int_{0}^{b} \left(\int_{x}^{\infty} \frac{\sin t}{t} \, dt \right) \, dx &amp;=  x \left( \int_{x}^{\infty} \frac{\sin t}{t} \, dt \right)\Bigg|^{b}_{0}+ \int_{0}^{b} \sin(x) \, dx \\ &amp;= b \int_{b}^{\infty} \frac{\sin t}{t} \, dt  -\cos(b) + 1 . \end{align}$$

But you showed that  $\int_{b}^{\infty} \frac{\sin t}{t} \, dt \sim \frac{\cos b}{b} + \frac{\sin b}{b^{2}} $ as $b \to \infty$.

So as $b \to \infty$, $b \int_{b}^{\infty} \frac{\sin t}{t} \, dt  -\cos(b)$ tends to zero, showing that the integral does indeed evaluate to $1$.

Strangely, Wolfram Alpha (unlike Mathematica) says that the integral doesn't converge.  But if you make the upper limit very large, it returns a value very close to $1$.
$\int_0^\infty \mathrm{Si}(x)\mathrm dx$
$\int_0^\infty \int_x^\infty \mathrm{\frac{sint}{t}}\mathrm dtdx$

Then by changing the order of integration we get

$\int_0^\infty \int_0^t \mathrm{\frac{sint}{t}}\mathrm dxdt$

$\int_0^\infty \mathrm{sinx}\mathrm dx$

Which diverges