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Finding a value of the following strange improper integral confronted me in the integral calculus working.

$$ \int_{0}^{\infty}\frac{\sin(x^{2})}{(x^{2}+1)^{3/2}}dx $$

I tried to bring this integral close to the Fresnel integral but it didn't work well.

How should I calculate? I have NO idea...

user
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  • A bit of context would not hurt. By using the Laplace transform, we may see that such integral depends on the values of the Bessel functions $J_0,J_1,Y_0,Y_1$ at $\frac{1}{2}$. – Jack D'Aurizio Sep 23 '15 at 15:37
  • @JackD'Aurizio how should I use the Laplace transform? I'm glad if you give bit more detail (particularly relationship to the Bessel functions). – user Sep 23 '15 at 15:49
  • @P.Mike: The irreducible presence of $x^2+1$ suggests a substitution of the form $x=\sinh t$. However, the only functions expressible as a definite integral whose integrand contains a non-trivial instance of $f\Big(g(x)\Big)$, where f and g are both trigonometric and/or hyperbolic functions, are the Bessel functions. See here for more information. – Lucian Sep 24 '15 at 06:22

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By substituting $x^2=z$ we have: $$ I = \frac{1}{2}\int_{0}^{+\infty}\frac{\sin(z)}{z^{1/2}(1+z)^{3/2}}\,dz\tag{1}$$ but since $\mathcal{L}\left(\sin(z)\right)=\frac{1}{s^2+1}$ and: $$ \mathcal{L}^{-1}\left(\frac{1}{z^{1/2}(1+z)^{3/2}}\right)= se^{-s/2}\left(I_0\left(\frac{s}{2}\right)-I_1\left(\frac{s}{2}\right)\right)\tag{2}$$ it follows that: $$ I = \frac{1}{2}\int_{0}^{+\infty}\frac{se^{-s/2}}{1+s^2}\left(I_0\left(\frac{s}{2}\right)-I_1\left(\frac{s}{2}\right)\right)\,ds\\=\frac{\pi}{4}\left[\left(J_1\left(\frac{1}{2}\right)-Y_0\left(\frac{1}{2}\right)\right)\cos\left(\frac{1}{2}\right)+\left(J_0\left(\frac{1}{2}\right)+Y_1\left(\frac{1}{2}\right)\right)\sin\left(\frac{1}{2}\right)\right].\tag{3} $$

Jack D'Aurizio
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  • Thank you for giving solution. I have one question. How did you calculate the inverse Laplace transform: $$ \mathcal{L}^{-1}\left(\frac{1}{z^{1/2}(1+z)^{3/2}}\right) $$ – user Sep 24 '15 at 01:53
  • Please let me ask one more question. On getting the solution (last line) from the integral (3), did you use some formula for integrals of $I_{0}$ and $I_{1}$? how did you calculate (3). – user Sep 24 '15 at 04:45