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!

  • 2
    $\begingroup$ 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. $\endgroup$ – Harald Hanche-Olsen Jan 10 '13 at 10:55
  • $\begingroup$ @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. $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – dan Jan 11 '13 at 9:29
  • $\begingroup$ @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. $\endgroup$ – Mhenni Benghorbal Jan 11 '13 at 12:41

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