# How to evaluate $\int \frac{\sin (\pi x)}{|x|^a + 1} dx$

How do I integrate this?

$$\int \frac{\sin (\pi x)}{|x|^a+1} \, dx$$

I really struggle to find a solution. I even tried Wolfram Alpha and Mathematica, but neither could give me an answer.

I have to find all a for that the Improper Integral exists. So my attempt is:

$$\lim_{r \to +\infty} \int_{-r}^r \frac{\text{Sin}[\pi x]}{1+\text{Abs}[x]^a} \, dx$$

Any ideas? Thanks

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Perhaps it cannot be expressed in terms of elementary functions (for general $a$). When $a$ is an integer it is not that difficult to integrate. –  glebovg Oct 22 '12 at 5:49
Indeed. Do you have any reason to think that there is a solution in the kind of form you want? –  Gerry Myerson Oct 22 '12 at 5:51
a is an integer. I have to find all a for that the Improper Integral exists. I edited the question above. –  kornnflake Oct 22 '12 at 5:58
Perhaps contour integration? Do a few cases and see if you can generalize and prove by induction. I do not think Mathematica can handle it, or you cannot even generalize it. –  glebovg Oct 22 '12 at 6:07
If you prefer Mathematica, consider a few cases $a = 1$, $a = 2$ and $a = 3$ (for example) and see if you can generalize. Then prove by induction or hope your generalization is correct. –  glebovg Oct 22 '12 at 6:11

## 2 Answers

You are only asked when the improper integral exists (that is, converges), not what the integral equals.

There are two senses in which the Improper Integral can exist:

$\;\text{1}$. Cauchy Principal Value: $\displaystyle\lim_{L\to\infty}\int_{-L}^L\frac{\sin(\pi x)}{|x|^a+1}\,\mathrm{d}x$

Since the integrand is odd, the integral is $0$ for any $L$, so the limit is $0$ for any $a$.

$\;\text{2}$. standard: $\displaystyle\lim_{L\to\infty}\int_0^L\frac{\sin(\pi x)}{|x|^a+1}\,\mathrm{d}x+\lim_{L\to\infty}\int_{-L}^0\frac{\sin(\pi x)}{|x|^a+1}\,\mathrm{d}x$

Since the integrand is odd, the integrals above are negatives of each other. Therefore, if one of the limits exists, both do.

Define $$b_n=(-1)^n\int_n^{n+1}\frac{\sin(\pi x)}{|x|^a+1}\,\mathrm{d}x=\int_n^{n+1}\frac{|\sin(\pi x)|}{|x|^a+1}\,\mathrm{d}x\lt\frac1{n^a+1}\tag{1}$$ then $$\sum_{k=0}^{n-1}(-1)^kb_k=\int_0^n\frac{\sin(\pi x)}{|x|^a+1}\,\mathrm{d}x\tag{2}$$ Note that \begin{align} b_n-b_{n+1} &=\int_n^{n+1}|\sin(\pi x)|\left(\frac{1}{|x|^a+1}-\frac{1}{|x+1|^a+1}\right)\,\mathrm{d}x\\ &\gtreqless0\text{ when }a\gtreqless0\tag{3} \end{align} When $a\le0$ the terms of the series in $(2)$ do not tend to $0$, so the series, and therefore the improper integral, does not converge.

When $a\gt0$, $b_n$ is a decreasing sequence, tending to $0$, and so by the Dirichlet Test, the series in $(2)$ converges. This handles the case for $L=n$, an integer. However, for $x\in[0,1]$ $$\int_n^{n+x}\frac{|\sin(\pi x)|}{|x|^a+1}\,\mathrm{d}x\lt\frac1{n^a+1}\tag{4}$$ thus the limit is true even when $L$ is not restricted to integers.

Summary

The Cauchy Principal Value of the improper integral exists for all $a$.

The standard improper integral exists only when $a>0$.

When the improper integral exists, its value is $0$ because the integrand is odd.

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Consider the two cases,

1) $a=2n,$ n a positive integer. In this case the integrand is an odd function, and the integaral

$$\int_{-r}^{r} \frac{\text{sin}(\pi x)}{1+x^{2n}} = 0 \implies \lim_{r\to \infty}\int_{-r}^{r} \frac{\text{sin}(\pi x)}{1+x^{2n}} = 0.$$

since the integrand is an odd function.

2) $a=2m+1,$ m is a positive integer. In this case split the interval of integration as

$$\int_{-r}^{r} \frac{\text{sin}(\pi x)}{1+|x|^{2m-1}}= \int_{-r}^{0} \frac{\text{sin}(\pi x)}{1-x^{2m-1}} + \int_{0}^{r} \frac{\text{sin}(\pi x)}{1+x^{2m-1}} \,.$$

Changing variables $x=-x$ in the first integral on RHS leads to

$$\int_{-r}^{r} \frac{\text{sin}(\pi x)}{1+|x|^{2m-1}}= -\int_{0}^{r} \frac{\text{sin}(\pi x)}{1+x^{2m-1}} + \int_{0}^{r} \frac{\text{sin}(\pi x)}{1+x^{2m-1}} =0 .$$

Taking the limit of the above equation as $r \to \infty$ follows the desired result.

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