TITLE: An almost Fresnel integral
QUESTION [14 upvotes]: $\def\d{\mathrm{d}}$So I tried doing the following integral:
$$I=\int_0^{+\infty}\sin(2^x)\,\d x,$$
which is quite similar to the famous Fresnel integral.  First, I rewrote $\sin$ using its complex exponential definition, then I let $u=2^x$:
$$I=\int_0^{+\infty}\frac{e^{i2^x}-e^{-i2^x}}{2i}\,\d x = \frac1{2i\ln(2)} \int_1^{+\infty} \frac{e^{iu}-e^{-iu}}u \,\d u.$$
(so close to letting me use Frullani's integral $\ddot\frown$)
But where do I go from here?  It looks very close to a place where I could use the exponential integral or something like that, but not quite...

REPLY [12 votes]: Hint. One may perform the change of  variable
$$
u=2^x, \quad \ln x= \frac1{\ln 2}\cdot \ln u, \quad dx=\frac1{\ln 2}\cdot \frac{du}u,
$$ giving
$$
I=\int_0^{+\infty}\sin(2^x)\ dx=\frac1{\ln 2}\cdot\int_1^{+\infty}\frac{\sin(u)}{u}\ du=\frac1{\ln 2}\cdot\left(\frac{\pi }{2}-\text{Si}(1)\right)
$$ where we have made use of the sine integral function $\text{Si}(\cdot)$.

REPLY [7 votes]: If you perform this change of variable:
$$y = 2^x \Rightarrow \begin{cases}
x = \frac{\log(y)}{\log(2)}\\
dx = \frac{dy}{y \log(2)}\\
x = 0 \Rightarrow y = 1\\
x = +\infty \Rightarrow y = +\infty
\end{cases},$$
then you obtain the following:
$$\int_0^{+\infty} \sin(2^x)dx = \frac{1}{\log(2)} \int_1^{+\infty} \frac{\sin(y)}{y} dy = \frac{\text{Si}(+\infty) - \text{Si}(1)}{\log(2)},$$
where $\text{Si}(x)$ is the sine integral. I guess you can't do more than this.