An almost Fresnel integral

Question

$\def\d{\mathrm{d}}$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...

Answer 1

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 $\text{Si}(\cdot)$.

Answer 2

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.