Evaluate $\int_{0}^{\infty} \frac {\cos x - 1}{x^{1+\alpha}}\ dx$, $0 < \alpha < 1$

Question

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.

Answer 1

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.

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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 $\alpha\in (0,2)$

$$\bbox[5px,border:2px solid #C0A000]{I(\alpha)=\frac{\pi}{2\Gamma(1+\alpha)\sin(\pi \alpha/2)}}\tag3\\$$

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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 $\alpha \in (0,2)$.

Answer 2

Integrating by parts gives that $$ \int_0^{+\infty}\frac{1-\cos(x)}{x^{1+\alpha}}dx=\frac{1}{\alpha}\int_0^{+\infty}\frac{\sin(x)}{x^{\alpha}}dx $$ Let $f(z):=\frac{e^{iz}}{z^{\alpha}}$ and let $\gamma_R$ be the contour integration $[0,R]\cup\left\{Re^{i\vartheta},\vartheta\in\left[0,\frac{\pi}{2}\right]\right\}\cup[iR,0]$. Then because $f$ has no singularities inside the contour, we have $$ \int_{\gamma_R}f(z)dz=0 $$ However, $$ \int_{\gamma_R}f(z)dz=\int_0^R f(t)dt+iR\int_0^{\frac{\pi}{2}}f(Re^{i\vartheta})e^{i\vartheta}d\vartheta-i\int_0^R f(it)dt $$ And, $$ \left|\int_0^{\frac{\pi}{2}}f(Re^{i\vartheta})e^{i\vartheta}d\vartheta\right|\leqslant\frac{1}{R^{\alpha}}\int_0^{\frac{\pi}{2}}e^{-R\sin\vartheta}d\vartheta\leqslant\frac{1}{R^{\alpha}}\int_0^{\frac{\pi}{2}}e^{-R\frac{2}{\pi}\vartheta}d\vartheta\ll\frac{1}{R^{1+\alpha}} $$ Therefore $$ \lim\limits_{R\rightarrow +\infty}iR\int_0^{\frac{\pi}{2}}f(Re^{i\vartheta})e^{i\vartheta}d\vartheta=0 $$ Taking the limit as $R\rightarrow +\infty$ gives that $$ \int_0^{+\infty}f(t)dt=i\int_0^{+\infty}f(it)dt $$ That is $$ \int_0^{+\infty}\frac{e^{it}}{t^{\alpha}}dt=e^{i\frac{\pi}{2}(1-\alpha)}\Gamma(1-\alpha) $$ Taking the imaginary part gives that $$ \int_0^{+\infty}\frac{\sin(t)}{t^{\alpha}}dt=\sin\left(\frac{\pi}{2}(1-\alpha)\right)\Gamma(1-\alpha)=-\cos\left(\frac{\pi\alpha}{2}\right)\alpha\Gamma(-\alpha) $$ We can therefore conclude that $$ \int_0^{+\infty}\frac{1-\cos(x)}{x^{1+\alpha}}dx=-\cos\left(\frac{\pi\alpha}{2}\right)\Gamma(-\alpha) $$ which is the formula given by Hans Olo in the comments section.