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In what conditions do these two improper integrals $$\int^\infty_0 x^{k} e^{P(x)} \sin (Q(x)) \mathrm{d}x \text{ and } \int^\infty_0 x^{k} e^{P(x)} \cos (Q(x)) \mathrm{d}x$$ converge, where $k$ is a real number, $P(x)$ is an $m$-degree polynomial with real number coefficients whose leading coefficient is positive, $Q(x)$ is an $n$-degree polynomial ($n\geq 2$) with real number coefficients?

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I changed $P$ to $Q$ at the end of the question. Please verify that this is correct. – Alex Becker May 9 '12 at 0:28
If my edit is correct, certainly $P$ must be constant. I believe the only other constraint is that $k\leq -1$. – Alex Becker May 9 '12 at 0:30
up vote 2 down vote accepted

The degree of $Q'$ is $n-1$, so there exist constants $C_1$ and $a_1$ such that $|Q'(x)| \le C x^{n-1}$ for $x\ge a_1$. Since $|Q(x)| \to \infty$ as $x\to\infty$, there exists a strictly increasing sequence $(x_j)$ with $x_j \to \infty$ and $\sin Q(x_j) = 1$. Let $\epsilon_j = \frac1{2 C x_j^{n-1}}$, and let $I_j = [x_j - \epsilon_j, x_j]$. For large $j$ we have that $x_j - \epsilon_j \ge a_1$, so $|Q'| \le Cx_j^{n-1}$ on $I_j$, implying that the image $J_j = Q(I_j)$ has length $\le \frac12$, so that $\sin y \ge \frac12$ for $y \in J_j$, i.e., that $\sin Q(x) \ge \frac12$ for $x\in I_j$.

Now if $P$ is non-constant with positive leading coefficient, then there exists constants $C_2>0$ and $a_2$ with $P(x) \ge C_2 x $ for $x \ge a_2$. Since $x^k e^{C_2\sqrt{x}} \to \infty$ as $x\to\infty$, for any value of $k$, there exists $a_3\ge a_2$ such that $x^k e^{P(x)} \ge x^k e^{C_2 x} \ge e^{\sqrt{x}}$ for $x \ge a_3$.

Taken together, this implies for large $j$: $$\int_{x_j-\epsilon_j}^{x_j} x^k e^{P(x)} \sin Q(x) \, dx \ge \frac12 \epsilon_j e^{\sqrt{x_j - \epsilon_j}} = \frac{e^{\sqrt{x_j - \epsilon_j}}}{4C x_j^{n-1}} \to \infty$$ as $j \to\infty$, so that the integral does not converge. The same argument shows that the other integral does not converge either if $P$ is non-constant with positive leading coefficient.

Now if $P$ is constant, the question is whether the integral $$ \int_0^\infty x^k \sin Q(x) \, dx $$ converges. For $k<0$ there is a singularity at $x=0$, which is always integrable if $-1<k<0$. For $-2<k\le -1$ it is integrable iff $\sin Q(0)=0$, for $-3<k \le -2$ it is integrable iff $\sin Q(0)=0$ and $Q'(0)=0$ etc. Similar results are true for the other integral, but since $\sin$ and $\cos$ are never simultaneously $0$, at least one of the integrals always diverges if $k\le -1$. For the convergence at $\infty$ one can observe that the sequence of zeros of $\sin Q(x)$ asymptotically looks like $x_j \approx c j^{1/n}$ with some constant $c>0$, and that $x_{j+1} - x_{j} \approx c((j+1)^{1/n} - j^{1/n}) \approx cj^{1/n}\frac{1}{nj}=\frac{c}{n} j^{1/n-1}$, so the integral near $\infty$ behaves like the alternating series $$ \sum_j (-1)^j (cj^{1/n})^k \frac{c}n j^{1/n-1} = C \sum_j (-1)^j j^{(k+1)/n-1}$$ with a constant $C$. This series converges iff the exponent is negative, i.e., iff $k+1 < n$. There are some details to be filled in, but I believe this can be made rigorous, i.e., both integrals converge iff $P$ is constant and $-1<k<n-1$, they both diverge if $k\ge n-1$, and one of them may converge if $k\le -1$ if $\sin Q(x)$ or $\cos Q(x)$ vanishes to the appropriate order.

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