Evaluate $\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy$

Question

How do I evaluate the following integral?

$$\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy=\frac{\pi}{\Gamma(1+\alpha)\sin(\frac{\pi\alpha}{2})}$$

Thank you in advance.

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I found the answer. Thank you all!

Answer 1

Let $I(\alpha)$, $0<\alpha<2$, be the integral given by

$$\begin{align} I(\alpha)&=\int_{-\infty}^\infty \frac{1-\cos ( y )}{|y|^{1+\alpha}}\,dy\\\\ &=2\int_0^\infty \frac{1-\cos ( y )}{y^{1+\alpha}}\,dy \tag 1 \end{align}$$

Integrating by parts $(1)$ with $u=1-\cos(y)$ and $v=-\frac{1}{\alpha y^{\alpha}}$ reveals

$$\begin{align} I(\alpha)&=\frac{2}{\alpha}\int_0^\infty \frac{\sin(y)}{y^\alpha}\,dy \tag 2\\\\ \end{align}$$

For $1<\alpha <2$, integrating by parts $(2)$ with $u=\sin(y)$ and $v=\frac{1}{1-\alpha}y^{1-\alpha}$ yields

$$\begin{align} I(\alpha)&=\frac{2}{\alpha(\alpha-1)}\int_0^\infty \frac{\cos(y)}{y^{\alpha-1}}\,dy \tag 3\\\\ \end{align}$$

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CASE $1$: $0<\alpha<1$

Note that for $0<\alpha<1$, we can write $(2)$ as

$$I(\alpha)=\frac{2}{\alpha}\text{Im}\left(\int_0^\infty \frac{e^{iy}}{y^\alpha}\,dy\right) \tag 4$$

Enforcing the substitution $y\to iy$ in the integral on the right-hand side of $(4)$ yields

$$\int_0^\infty \frac{e^{iy}}{y^\alpha}\,dy=e^{i\pi (1-\alpha)/2}\int_0^{-i\infty} \frac{e^{-y}}{y^\alpha}\,dy \tag 5$$

Using Cauchy's Integral Theorem, we can deform the contour back to the real line and write $(5)$ as

$$\begin{align} \int_0^\infty \frac{e^{iy}}{y^\alpha}\,dy&=e^{i\pi (1-\alpha)/2}\int_0^\infty y^{-\alpha}e^{-y}\,dy\\\\&=e^{i\pi (1-\alpha)/2}\Gamma(1-\alpha) \tag 6 \end{align}$$

Substituting $(6)$ into $(4)$, we obtain

$$\begin{align} \int_{-\infty}^\infty \frac{1-\cos ( y )}{|y|^{1+\alpha}}\,dy&=\frac{2}{\alpha}\sin((1-\alpha)\pi/2)\Gamma(1-\alpha)\\\\ &=\frac{2}{\alpha}\cos(\pi \alpha/2)\Gamma(1-\alpha) \tag 7\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\pi}{\Gamma(1+\alpha)\sin(\pi \alpha/2)}} \tag 8 \end{align}$$

where in going from $(7)$ to $(8)$ we used the functional relationship $\Gamma(1+z)=z\Gamma(z)$ along with Euler's Reflection formula $\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}$

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CASE $2$: $1<\alpha<2$

Note that for $1<\alpha<2$, we can write $(3)$ as

$$I(\alpha)=\frac{2}{\alpha(\alpha -1)}\text{Re}\left(\int_0^\infty \frac{e^{iy}}{y^{\alpha-1}}\,dy\right) \tag 9$$

Enforcing the substitution $y\to iy$ in the integral on the right-hand side of $(9)$ yields

$$\int_0^\infty \frac{e^{iy}}{y^{\alpha-1}}\,dy=-e^{-i\pi \alpha/2}\int_0^{-i\infty} \frac{e^{-y}}{y^{\alpha-1}}\,dy \tag {10}$$

Using Cauchy's Integral Theorem, we can deform the contour back to the real line and write $(10)$ as

$$\begin{align} \int_0^\infty \frac{e^{iy}}{y^{\alpha-1}}\,dy&=-e^{i\pi \alpha/2}\int_0^\infty y^{1-\alpha}e^{-y}\,dy\\\\&=-e^{i\pi \alpha/2}\Gamma(2-\alpha) \tag{11} \end{align}$$

Substituting $(11)$ into $(9)$, we obtain

$$\begin{align} \int_{-\infty}^\infty \frac{1-\cos ( y )}{|y|^{1+\alpha}}\,dy&=\frac{2}{\alpha(\alpha-1)}\cos(\alpha\pi/2)\Gamma(2-\alpha) \tag {12}\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\pi}{\Gamma(1+\alpha)\sin(\pi \alpha/2)}}\tag {13} \end{align}$$

where in going from $(12)$ to $(13)$ we used the functional relationship $\Gamma(1+z)=z\Gamma(z)$ along with Euler's Reflection formula $\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}$.

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PUTTING IT ALL TOGETHER:

Using $(8)$ and $(13)$ along with the well-known result $I(1)=\pi/2$, we find that for all $0<\alpha<2$ we have

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \frac{1-\cos ( y )}{|y|^{1+\alpha}}\,dy=\frac{\pi}{\Gamma(1+\alpha)\sin(\pi \alpha/2)}}$$

as was to be shown!

Answer 2

Just an idea:

$$ \int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy = 2\int_0^{\infty}{\sum_{k=0}^{\infty}{ \frac{{\left(y\right)}^{2k+1-\alpha}}{(2k)!} } dy}$$

Let me know if it helps any.

Its complex integration according to one your tags....do you care what method is used for it?

Answer 3

Thank you Dr.MV for your answer. Since you found the solution first, I voted your answer as the correct one.

I was also able to find the solution, though in a different way. So for the sake of completeness, I´ll live it here.

First, let´s rewrite the integral as $$\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy=2\int_{0}^{\infty} \frac{(1-\cos { y } )}{y^{1+\alpha}}dy$$ Then use the folowing integral representation: $$\frac{1}{y^{1+\alpha}}=\frac{1}{\Gamma(1+\alpha)}\int_{0}^{\infty}u^{\alpha}e^{-yu} du$$ to rewrite the original integral as a double integral

$$\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy=\frac{2}{\Gamma(1+\alpha)}\int_{0}^{\infty}\int_{0}^{\infty} u^{\alpha}e^{-yu}(1- \cos{y}) dy du$$

Swap the order of integration and then concentrate in the inner integral first, which can be splited into two intergals

$$\int_{0}^{\infty}e^{-yu}dy - \int_{0}^{\infty} e^{-yu}\cos{y}dy$$

the result is:

$$\frac{1}{u}-\frac{u}{u^{2}+1} $$

plugging it back into the double integral leads us to:

$$\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy=\frac{2}{\Gamma(1+\alpha)}\int_{0}^{\infty} u^{\alpha} \left[\frac{1}{u}-\frac{u}{u^{2}+1}\right]du$$

$$=\frac{2}{\Gamma(1+\alpha)}\int_{0}^{\infty} \frac{u^{\alpha}}{u(u^{2}+1)}du =\frac{2}{\Gamma(1+\alpha)}\int_{0}^{\infty} \frac{u^{\alpha-1}}{(u^{2}+1)}du$$

now, make the substitution $$u=v^{2} \Rightarrow du=\frac{1}{2}v^{-\frac{1}{2}}dv$$

$$\frac{2}{\Gamma(1+\alpha)}\int_{0}^{\infty} \frac{v^{\frac{\alpha-2}{2}}}{(v+1)}\frac{dv}{2}$$

using the complex analysis result:

$$\int_{0}^{\infty} \frac{x^{\alpha-1}}{(x+1)}dx=\frac{\pi}{\sin{\pi\alpha}}$$

Leads us to the final answer

$$\int_{-\infty}^{\infty} \frac{(1-\cos { y } )}{\mid{y}\mid^{1+\alpha}}dy=\frac{\pi}{\Gamma(1+\alpha)\sin(\frac{\pi\alpha}{2})}$$