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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – glebovg
    Oct 22, 2012 at 5:49
  • $\begingroup$ Indeed. Do you have any reason to think that there is a solution in the kind of form you want? $\endgroup$ Oct 22, 2012 at 5:51
  • $\begingroup$ a is an integer. I have to find all a for that the Improper Integral exists. I edited the question above. $\endgroup$
    – krnflake
    Oct 22, 2012 at 5:58
  • $\begingroup$ 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. $\endgroup$
    – glebovg
    Oct 22, 2012 at 6:07
  • $\begingroup$ 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. $\endgroup$
    – glebovg
    Oct 22, 2012 at 6:11

2 Answers 2


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.


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.


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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