relative of cosine integral

For determining of the value of the improper integral

$$I(a):={\int }_0^{\mathrm{\infty }}\frac{\cos a{x}^2-\cos ax}{x}dx\mathit{\hspace{1em}\hspace{1em}}(a>0),$$

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):

$$ℒ\{I(t)\}=ℒ\{{\int }_0^{\mathrm{\infty }}(\cos t{x}^2-\cos tx)\frac{dx}{x}\}={\int }_0^{\mathrm{\infty }}\left(\frac{s}{s^2+x^4}-\frac{s}{s^2+x^2}\right)\frac{dx}{x}$$

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

$ℒ\{I(t)\}$$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{s}}{\displaystyle {\int }_0^{\mathrm{\infty }}}\left({\displaystyle \frac{1}{x}}-{\displaystyle \frac{x^3}{s^2+x^4}}-{\displaystyle \frac{1}{x}}+{\displaystyle \frac{x}{s^2+x^2}}\right)𝑑x$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{s}}\underset{x=0}{\overset{\mathrm{\infty }}{/}}\left[{\displaystyle \frac{1}{2}}\ln (s^2+x^2)-{\displaystyle \frac{1}{4}}\ln (s^2+x^4)\right]$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{4}}\underset{x=0}{\overset{\mathrm{\infty }}{/}}\ln {\displaystyle \frac{{(s^2+x^2)}^2}{s^2+x^4}}=-{\displaystyle \frac{\ln s}{2s}}.$

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.  $ℒ\{1\}={\displaystyle \frac{1}{s}}$), and thus we have the result

$$I(a)=\frac{\gamma +\ln a}{2}.$$