How to evaluate $\int_0^\infty \frac{\cos(ax)}{1+x^N}dx$ for $a,N\in\mathbb{R}$ and $N> 1$

Solution 1:

For even $N$ you can use the residue theorem as well. It suffices to consider $a>0$ and calculate the integral $$\oint_{-\infty}^{\infty} \frac{e^{iax}}{1+x^N} \, {\rm d}x = \int_{0}^{\infty} \frac{2\cos(ax)}{1+x^N} \, {\rm d}x = 2\pi i \sum_{0 \leq k \leq \frac{N-1}{2}} {\rm Res}\left( \frac{e^{iax}}{1+x^N} \right)\Bigg|_{x=e^{\frac{i\pi+i2\pi k}{N}}} \\ =\frac{-2\pi i}{N} \sum_{k=0}^{N/2-1} e^{ia e^{\frac{i\pi+i2\pi k}{N}} + \frac{i\pi+i2\pi k}{N}} = \frac{-2\pi i}{N} \sum_{k=0}^{N/2-1} e^{ia \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{i\pi+i2\pi k}{N}}e^{-a\sin\left( \frac{\pi+2\pi k}{N} \right)} \\ =\frac{-2\pi i}{N} \sum_{k=0}^{N/2-1} \Bigg[\cos\left(a \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{\pi+2\pi k}{N}\right) \\ + i \sin\left(a \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{\pi+2\pi k}{N}\right)\Bigg]e^{-a\sin\left( \frac{\pi+2\pi k}{N} \right)} \\ =\frac{2\pi}{N} \sum_{k=0}^{N/2-1} \sin\left(a \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{\pi+2\pi k}{N}\right)e^{-a\sin\left( \frac{\pi+2\pi k}{N} \right)}$$ where the loop-integral on the LHS is closed in the upper half plane. The imaginary part is zero, since the LHS is real.

Solution 2:

Just some working out in progress.

Well, we are trying to find the following integral:

$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right):=\int_0^\infty\frac{\cos\left(\text{k}x\right)}{\alpha^2+x^\text{n}}\space\text{d}x\tag1$$

To calculate the integral, we first note we may write that:

$$\frac{1}{\alpha^2+x^\text{n}}=\int_0^\infty\exp\left(-\left(\alpha^2+x^\text{n}\right)\text{y}\right)\space\text{dy}\tag2$$

And then we have:

$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right)=\int_0^\infty\left(\int_0^\infty\cos\left(\text{k}x\right)\exp\left(-\left(\alpha^2+x^\text{n}\right)\text{y}\right)\space\text{dy}\right)\space\text{d}x\tag3$$

Change the integration order assuming the value of the integral preserves:

$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right)=\int_0^\infty\exp\left(-\text{y}\alpha^2\right)\left(\int_0^\infty\cos\left(\text{k}x\right)\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x\right)\space\text{dy}\tag4$$

Make use of the power series of $\cos x$:

$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right)=\int_0^\infty\exp\left(-\text{y}\alpha^2\right)\left(\int_0^\infty\sum_{\text{m}\space\ge\space0}\frac{\left(-1\right)^\text{m}\left(\text{k}x\right)^{2\text{m}}}{\left(2\text{m}\right)!}\cdot\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x\right)\space\text{dy}\tag5$$

Which gives:

$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right)=\int_0^\infty\exp\left(-\text{y}\alpha^2\right)\left(\sum_{\text{m}\space\ge\space0}\frac{\left(-1\right)^\text{m}\text{k}^{2\text{m}}}{\left(2\text{m}\right)!}\int_0^\infty x^{2\text{m}}\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x\right)\space\text{dy}\tag6$$

If you substitute:

$$\text{s}=\text{y}^\frac{2\text{m}+1}{\text{n}}\cdot x^{2\text{m}+1}\tag7$$

We will get:

$$\int_0^\infty x^{2\text{m}}\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x=\frac{\text{y}^{-\left(\frac{2\text{m}}{\text{n}}+\frac{1}{\text{n}}\right)}}{2\text{m}+1}\int_0^\infty\exp\left(-\text{s}^\frac{\text{n}}{2\text{m}+1}\right)\space\text{ds}\tag8$$

Which is defined in terms of the incomplete gamma function:

$$\int_0^\infty\exp\left(-\text{s}^\frac{\text{n}}{2\text{m}+1}\right)\space\text{ds}=-\frac{2\text{m}+1}{\text{n}}\cdot\left[\Gamma\left(\frac{2\text{m}+1}{\text{n}},\text{s}^\frac{\text{n}}{2\text{m}+1}\right)\right]_0^\infty\tag9$$

So:

$$\int_0^\infty x^{2\text{m}}\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x=\frac{\text{y}^{-\left(\frac{2\text{m}}{\text{n}}+\frac{1}{\text{n}}\right)}}{\text{n}}\cdot\left[\Gamma\left(\frac{2\text{m}+1}{\text{n}},\text{s}^\frac{\text{n}}{2\text{m}+1}\right)\right]_\infty^0\tag{10}$$

Which gives:

$$\int_0^\infty x^{2\text{m}}\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x=\frac{\text{y}^{-\left(\frac{2\text{m}}{\text{n}}+\frac{1}{\text{n}}\right)}}{\text{n}}\cdot\Gamma\left(\frac{2\text{m}+1}{\text{n}}\right)\tag{11}$$

So:

$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right)=\frac{1}{\text{n}}\sum_{\text{m}\space\ge\space0}\frac{\left(-1\right)^\text{m}\text{k}^{2\text{m}}}{\left(2\text{m}\right)!}\cdot\Gamma\left(\frac{2\text{m}+1}{\text{n}}\right)\int_0^\infty\exp\left(-\text{y}\alpha^2\right)\text{y}^{-\left(\frac{2\text{m}}{\text{n}}+\frac{1}{\text{n}}\right)}\space\text{dy}\tag{12}$$