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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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2  
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 2

up vote 1 down vote accepted

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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