Definite integral with $\mathrm{Si}$ in integrand

Question

Does the function

$$f(t) = \int_0^{\sqrt{3}} (x^2-1) \;\mathrm{Si}((x^2-1)\,t)\; \mathrm{d}x$$

have a representation in terms of elementary functions of $t$ for real, positive $t$? Here, $\mathrm{Si}(x)$ is the sine integral:

$$\mathrm{Si}(x) = \int_0^x \frac{\sin(u)}{u} \mathrm{d}u = \int_0^1 \frac{\sin(u x)}{u} \mathrm{d}u.$$

I have tried several variable transformations and integral representations of the sine integral followed by interchanging the order of integration (see e.g. http://functions.wolfram.com/GammaBetaErf/SinIntegral/). This did not work out (but could be limited by my dated math capabilities). A reasonably fast converging power series representation will also do.

Thank you!

Answer 1

Here is a start, using integration by parts with $u=\mathrm{Si}((x^2-1)\,t)$, we have

$$f(t) = \int_0^{\sqrt{3}} (x^2-1) \;\mathrm{Si}((x^2-1)\,t)\; \mathrm{d}x= -\frac{2}{3}\,\int _{0}^{\sqrt {3}}\!{\frac { \left( {x}^{2}-3 \right) {x}^{2} \sin \left( \left( {x}^{2}-1 \right) t \right) }{{x}^{2}-1}}{dx} $$

$$ = -\frac{2}{3}\,\int _{0}^{\sqrt {3}}\!{ { {x}^{2} \sin\left(\left( {x}^{2}-1 \right) t \right) }}{dx}+\frac{4}{3}\,\int _{0}^{\sqrt {3}}\!{ { \frac{{x}^{2}}{x^2-1} \sin\left(\left( {x}^{2}-1 \right) t \right) }}{dx} = \dots. $$

Added: For the first integral, maple was able to give the answer

$$-\frac{2}{3}\,\int _{0}^{\sqrt {3}}\!{ { {x}^{2} \sin\left(\left( {x}^{2}- 1 \right) t \right)}}{dx}=\frac{2\sqrt {3}}{3}\,{\frac{\left( \cos\left(t\right)\right)^{2}}{t}} -\frac{\sqrt {3}}{3}\,{\frac {1 }{t}}$$ $$-\frac{1}{\sqrt{3}}\,{\frac {\cos \left(t\right)\, {_1F_2(\frac{1}{4};\,\frac{1}{2},\frac{5}{4};\,-\frac{9}{4}\,{t}^{2})}}{t}}$$ $$-\frac{1}{\sqrt{3}}\,\sin\left(t \right)\, {_1F_2(\frac{3}{4};\,\frac{3}{2},\frac{7}{4};\,-\frac{9}{4}\,{t}^{2})} ,$$

where $_1F_2$ is the hypergeometric function.

Note: I paid attention to Olsen's suggestion after I had posted my answer. I already upvoted.