# relative of cosine integral

For determining of the value of the improper integral

 $I(a)\;:=\;\int_{0}^{\infty}\frac{\cos{ax^{2}}-\cos{ax}}{x}\,dx\qquad(a>0),$

related to the cosine integral, we think it as a function of the parametre $a$ which we denote by $t$.  Then we can take the Laplace transform (see the integration with respect to a parametre in the table of Laplace transforms):

 $\mathcal{L}\{I(t)\}\;=\;\mathcal{L}\{\int_{0}^{\infty}(\cos{tx^{2}}-\cos{tx})% \frac{dx}{x}\}\;=\;\int_{0}^{\infty}\left(\frac{s}{s^{2}\!+\!x^{4}}-\frac{s}{s% ^{2}\!+\!x^{2}}\right)\frac{dx}{x}$

Splitting the fractional expressions to http://planetmath.org/node/5812partial fractions and integrating give

 $\displaystyle\mathcal{L}\{I(t)\}$ $\displaystyle\;=\;\frac{1}{s}\int_{0}^{\infty}\left(\frac{1}{x}-\frac{x^{3}}{s% ^{2}\!+\!x^{4}}-\frac{1}{x}+\frac{x}{s^{2}\!+\!x^{2}}\right)dx$ $\displaystyle\;=\;\frac{1}{s}\!\operatornamewithlimits{\Big{/}}_{\!\!\!x=0}^{% \,\quad\infty}\left[\frac{1}{2}\ln(s^{2}\!+\!x^{2})-\frac{1}{4}\ln(s^{2}\!+\!x% ^{4})\right]$ $\displaystyle\;=\;\frac{1}{4}\!\operatornamewithlimits{\Big{/}}_{\!\!\!x=0}^{% \,\quad\infty}\ln\frac{(s^{2}\!+\!x^{2})^{2}}{s^{2}\!+\!x^{4}}\;\,=\;\,-\frac{% \ln{s}}{2s}.$

As seen in the http://planetmath.org/node/10588table of Laplace transforms, the gotten expression is the Laplace transform of  $\displaystyle\frac{\gamma+\ln{t}}{2}\,=\,I(t)$  (N.B.  $\displaystyle\mathcal{L}\{1\}=\frac{1}{s}$), and thus we have the result

 $I(a)\;=\;\frac{\gamma+\ln{a}}{2}.$
Title relative of cosine integral RelativeOfCosineIntegral 2013-03-22 18:44:35 2013-03-22 18:44:35 pahio (2872) pahio (2872) 8 pahio (2872) Example msc 44A10 SubstitutionNotation EulersConstant RelativeOfExponentialIntegral IntegrationOfLaplaceTransformWithRespectToParameter