Define $\mathrm{si}(x) := \displaystyle\int_x^\infty \frac{\sin(t)}{t}\mathrm{d}t $ for all $x>0$. 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 $$ \mathrm{si}(x) = \frac{\cos(x)}x + \frac{\sin(x)}{x^2} + O\left(\frac{1}{x^2}\right) = \frac{\cos(x)}x + O\left(\frac{1}{x^2}\right). $$

  • $\begingroup$ Due to Mathematica, it's $1$. $\endgroup$ – Friedrich Philipp Mar 6 '16 at 22:26
  • $\begingroup$ What about a manual proof ? $\endgroup$ – Groovy. Mar 6 '16 at 22:29
  • $\begingroup$ BTW, how do you show by integration by parts that $(\sin t)/t$ is integrable over $\mathbb R^+$? $\endgroup$ – Friedrich Philipp Mar 6 '16 at 22:34
  • $\begingroup$ I meant for $\mathrm{Si}$, but I'll show both: let $x$ be a strictly positive real number. Then, $\int_1^x \sin(t)/t \mathrm d t = \cos(x)/x - \cos(1) - \int_1^x \cos(t)/t^2\mathrm d t$. The second integral converges, the first terms has limit $0$ as $x\to \infty$, thus the integral of $\sin(t)/t$ converges. $\endgroup$ – Groovy. Mar 6 '16 at 22:37
  • $\begingroup$ Well, $\int_{-x}^x t\,dt = 0$. Is $t\mapsto t$ integrable over $\mathbb R$? You have to show that the absolute value of the integrand is integrable, which means that you will have to split the integral. $\endgroup$ – Friedrich Philipp Mar 6 '16 at 22:42

Integrating by parts, $$ \begin{align} \int_{0}^{b} \left(\int_{x}^{\infty} \frac{\sin t}{t} \, dt \right) \, dx &= x \left( \int_{x}^{\infty} \frac{\sin t}{t} \, dt \right)\Bigg|^{b}_{0}+ \int_{0}^{b} \sin(x) \, dx \\ &= b \int_{b}^{\infty} \frac{\sin t}{t} \, dt -\cos(b) + 1 . \end{align}$$

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

  • 1
    $\begingroup$ Does the function verify the conditions for Fubini's theorem ? $\endgroup$ – Groovy. Mar 6 '16 at 22:55

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