I am interested in finding a formula for the evaluation of
\begin{equation}
\int_0^\infty \frac{\cos(x) – 1}{x^{1+\alpha}}dx, \quad 0<\alpha<1.
\end{equation}
I believe the integral exists as an improper integral, however the calculation becomes tricky due to our choice of $\alpha$. Given the similar look to integrating $\sin x /x$, I decided to try using the Laplace transform. So I let
\begin{equation}
A(t) = \int_0^\infty \frac{\cos(tx) – 1}{x^{1+\alpha}}dx
\end{equation}
and then do
\begin{align}
\mathcal{L}(A)(t) ={}& \int_0^\infty \int_0^\infty \frac{\cos(tx) – 1}{x^{1+\alpha}} e^{-st}dx dt\\
={}& \int_0^\infty \frac{1}{x^{1+\alpha}} \int_0^\infty (\cos(tx) – 1)e^{-st}dtdx\\
={}& \int_0^\infty \frac{1}{x^{1+\alpha}} \mathcal{L}(\cos(tx) – 1)(t)dx\\
={}& \int_0^\infty \frac{1}{x^{1+\alpha}} \left( \frac{s}{s^2+x^2} – \frac{1}{s} \right)dx\\
={}& – \frac{1}{s} \int_0^\infty \frac{x^{1-\alpha}}{s^2+x^2}dx.
\end{align}
Unfortunately I've found myself unable to solve this improper integral. Given that our only condition on $\alpha$ is $0 < \alpha < 1$, it doesn't seem to lend itself to trigonometric substitution. Integration by parts didn't appear to clean anything up either.
I greatly appreciate any clarity or ideas. To be honest, even if I can solve the last improper integral, I can't be sure the inverse Laplace transform will be clean either; so perhaps my initial approach is where I need work. Thank you.
Best Answer
I thought it might be instructive to present an approach that circumvents use of complex analysis and relies on real analysis only. To that end, we now proceed.
Let $I(\alpha)$, $0<\alpha<2$, be given by the integral
$$I(\alpha)=\int_0^\infty \frac{1-\cos(x)}{x^{1+\alpha}}\,dx$$
Let $f(t)=1-\cos(t)$ and $G(s) = \frac1{s^{1+\alpha}}$. Then appealing to this property of the Laplace Transform, we have
$$\begin{align} I(\alpha)&=\int_0^\infty \color{blue}{\mathscr{L}\{f\}(x)} \color{red}{ \mathscr{L^{-1}}\{G\}(x)}\,dx\\\\ &=\int_0^\infty \color{blue}{\frac{1}{x(x^2+1)}}\,\,\,\color{red}{\frac{x^{\alpha}}{\Gamma(1+\alpha)}}\,dx\\\\ &=\frac{1}{\Gamma(1+\alpha)}\int_0^\infty \frac{x^{\alpha-1}}{x^2+1}\,dx\tag1 \end{align}$$
In THIS ANSWER, I showed using real analysis only that the integral on the right-hand side of $(1)$ is given by
$$\int_0^\infty \frac{x^{\alpha-1}}{x^2+1}\,dx=\frac{\pi}{2}\sec\left(\frac{\pi}{2}(\alpha-1)\right)\tag2$$
Using $(2)$ in $(1)$ yields for $\alpha\in (0,2)$
$$\bbox[5px,border:2px solid #C0A000]{I(\alpha)=\frac{\pi}{2\Gamma(1+\alpha)\sin(\pi \alpha/2)}}\tag3\\$$
NOTE: ALTERNATIVE FORM
Using the reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ for the Gamma function, we can write $(3)$ alternatively as for $\alpha\in (0,2)\setminus { 1}$
$$\bbox[5px,border:2px solid #C0A000]{I(\alpha)=-\cos(\pi \alpha)\Gamma(-\alpha)}\tag4$$
which agrees with previously reported results.
For $\alpha\to 1$, the right-hand side of $(4)$ approaches $\pi/2$. So, if we define $(4)$ as a function with a removeable discontinuity at $\alpha=1$, the the result holds for $\alpha \in (0,2)$.