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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. 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!

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Have you tried a partial integration to replace the Si term with its derivative? Or possibly, just finding $f'(t)$ by differentiating under the integral sign. – Harald Hanche-Olsen Jan 10 '13 at 10:55
@HaraldHanche-Olsen: Interesting suggestion. Indeed, differentiating under the integral with respect to $t$, evaluating the integral over $x$, and integrating over $t$ (using the definitions of the Fresnel integrals that occur in the process), I get an expression regenerating the thought integral plus this one: $\int_0^{\sqrt{3}} \mathrm{Si}(t(1-y^2))\mathrm{d}y$ If I found a solution to this one, I could solve for the original integral. – dan Jan 10 '13 at 12:50

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.

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Good start. I am wondering about the meaning of your ellipsis. Do you imply that your second integral is straight-forward? If this is indeed true, please give me another hint. I have changed variables to yield, $\frac{1}{2}\int_{-1}^2\frac{\sqrt{y+1} sin(yt)}{y}\mathrm{d}y$. However, now I am stuck. Expanding $\sqrt{y+1}$ using the binomial theorem, I tried to obtain a series representation. Alternatives? – dan Jan 11 '13 at 9:29
@dan: Try to use the power series of $\sin(yt)$ and manipulate the resulting integral. I have not worked on it and I'll have a look at it later. – Mhenni Benghorbal Jan 11 '13 at 12:41

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