integration - Find $int_{0}^{infty }frac{cos x-cos x^2}{x}mathrm dx$

Recently, I met a integration below

\begin{align*}
\int_{0}^{\infty }\frac{\sin x-\sin x^2}{x}\mathrm{d}x&=\int_{0}^{\infty }\frac{\sin x}{x}\mathrm{d}x-\int_{0}^{\infty }\frac{\sin x^{2}}{x}\mathrm{d}x\\
&=\int_{0}^{\infty }\frac{\sin x}{x}\mathrm{d}x-\frac{1}{2}\int_{0}^{\infty }\frac{\sin x}{x}\mathrm{d}x\\
&=\frac{1}{2}\int_{0}^{\infty }\frac{\sin x}{x}\mathrm{d}x=\frac{\pi }{4}
\end{align*}
the same way seems doesn't work in
$$\int_{0}^{\infty }\frac{\cos x-\cos x^2}{x}\mathrm dx$$
but why? Then how to evaluate it? Thx!

Answer

Because $\displaystyle \int_{0}^{\infty }\frac{\cos x}{x}\, \mathrm{d}x$ does not converge, you can see here for a proof.

So we have to find another way to evaluate it.

I'll think about it and post a solution later.

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Solution:
\begin{align*}
\int_{0}^{\infty }\frac{\cos x-\cos x^2}{x}\,\mathrm{d}x&=\lim_{\alpha \rightarrow \infty }\int_{0}^{\alpha }\frac{\cos x-\cos x^2}{x}\,\mathrm{d}x=\lim_{\alpha \rightarrow \infty }-\int_{0}^{\alpha }\frac{1-\cos x+\cos x^2-1}{x}\,\mathrm{d}x\\

&=\lim_{\alpha \rightarrow \infty }\left ( -\int_{0}^{\alpha }\frac{1-\cos x}{x}\,\mathrm{d}x+\int_{0}^{\alpha }\frac{1-\cos x^2}{x}\,\mathrm{d}x \right )\\
&=\lim_{\alpha \rightarrow \infty }\left ( -\int_{0}^{\alpha }\frac{1-\cos x}{x}\,\mathrm{d}x+\frac{1}{2}\int_{0}^{\alpha^{2} }\frac{1-\cos x}{x}\,\mathrm{d}x \right )\\
&=\lim_{\alpha \rightarrow \infty }\left \{ \mathrm{Ci}\left ( \alpha \right )-\gamma -\ln\alpha +\frac{1}{2}\left [ \gamma +\ln\alpha ^{2}-\mathrm{Ci}\left ( \alpha ^{2} \right ) \right ] \right \}\\
&=\lim_{\alpha \rightarrow \infty }\left [ -\frac{\gamma }{2}+\mathrm{Ci}\left ( \alpha \right )-\frac{1}{2}\mathrm{Ci}\left ( \alpha ^{2} \right ) \right ]\\
&=-\frac{\gamma }{2}
\end{align*}
where $\mathrm{Ci}\left ( \cdot \right )$ is Cosine Integral and we can easily find that $\mathrm{Ci}\left ( \alpha \right )$ goes to $0$ when $\alpha \rightarrow \infty $.