Study the convergence of the following integrals:

(1). $\dps{\int_{-\infty}^{+\infty} x^ne^{-\sex{x^2+\frac{1}{x^2}}}\rd x}$ ($n $ for natural number).

(2). $\dps{\int_0^{+\infty} \sin^2\sez{\pi\sex{x+\frac{1}{x}}}\rd x}$.

Answer:

(1). $$\bex \int_{-\infty}^{+\infty}x^ne^{-\sex{x^2+\frac{1}{x^2}}}\rd x =\int_{-\infty}^{-1}+\int_{-1}^0 +\int_0^1 +\int_1^{+\infty} x^ne^{-\sex{x^2+\frac{1}{x^2}}}\rd x =i_1+i_2+i_3+i_4. \eex$$ to $I _1,i_4$, by $$\bex |x^ne^{-\sex{x^2+\frac{1}{x^2}}}| \leq |x|^n e^{-|x|^2}, \quad \lim_{|x|\to \infty} \frac{|x|^ne^{-|x|^2}}{1/|x|^2} =\lim_{t\to+\infty} \frac{t^{n+2}}{e ^{t^2}}=0 \eex$$ and comparative discriminant method are known $I _1,i_4$ absolute convergence. For $I _2,i_3$, the $$\bex \vlm{x} x^ne^{-\sex{x^2+\frac{1}{x^2}}}=0 \eex$$ is defined by the extension of the integrand $x =0$ and then successively on the $[-1,1]$. In conclusion, the original anomalous integral is absolutely convergent.

(2). Make $$\bex f (x) =x+\frac{1}{x},\quad f^{-1} (y) =\frac{y+\sqrt{y^2-4}}{2},\quad x\geq 1,\ Y\geq 2. \eex$$ to $\forall\ k\in\bbn$, take $$\bex a_k=f^{-1}\sex{2k+\frac{1}{4}},\quad b_k=f^{-1}\sex{2k+\frac{3}{4}, \eex$$ have $$ \bex A_k\leq X\leq B_k\ra 2k+\frac{1}{4}\leq f (x) =x+\frac{1}{x}\leq 2k+\frac{3}{4} \ra \sin^2\sez{\pi\sex{x+\frac{1}{x }}}\geq \frac{1}{2}, \eex$$ $$\beex \bea \int_{a_k}^{b_k}\sin^2\sez{\pi\sex{x+\frac{1}{x}}}\rd x &\geq \frac{1}{2} (B_k-a_k) \ &=\frac{1}{2}\sez{f^{-1} (Z_k)-f^{-1} (y_k)}\quad\sex{z_k=2k+\frac{3}{4},\ y_k=2k+\frac{1}{4}}\\ &=\frac{1}{4}\sez{z_k-y_k+\sqrt{z^2-4}-\sqrt{y_k^2-4}}\\ &\geq \frac{1}{4} (z_k-y_k) \ &= \frac{1}{8}. \eea \eeex$$ by the $A _k\to +\infty\ (k\to\infty) $ and Cauchy convergence criterion is known as the original integral divergence.

[Typical problems and methods in mathematical analysis of Periven exercises reference Solutions]4.5.7