How can I see if this integral is convergent or not $\int_0^\infty \ \frac{1}{1 + x^4\sin x} \,dx $

Question

I think the integral is convergent, but I don't know how to prove it.

$\int_0^\infty \ \frac{1}{1 + x^4\sin x} \,dx $

Answer 1

Let $f(x)=1+x^4\sin x$. Its first positive zero $x_0\simeq 3.1517$ is slightly greater than $\pi$. At this point, $f'(x_0)=4x_0^3\sin x_0+x_0^4\cos x_0\simeq -99.9362\neq 0$. So $f(x)\sim f'(x_0)(x-x_0)$ whence $$ \frac{1}{1+x^4\sin x}=\frac{1}{f(x)}\sim\frac{1}{f'(x_0)(x-x_0)} $$ when $x$ approaches $x_0$. So the integrand is continuous and positive on $[0,x_0)$, and by comparison $$ \int_0^{x_0}\frac{1}{1+x^4\sin x}dx=+\infty. $$ A fortiori, this function is not integrable over $[0,+\infty)$.

Answer 2

The integrand has an order $1$ singularity at each root of $1+x^4\sin(x)$. The real roots are located, for $n\ge1$, on the real axis at $$ x_n\approx n\pi-\frac{(-1)^n}{n^4\pi^4}\tag{1} $$ There is one simple root at $x_0=0.79474586313959135719 + 0.61883482670901662817 i$.

The residue at $x_n$ is $$ \frac1{4x_n^3\sin(x_n)+x_n^4\cos(x_n)}\approx\frac{(-1)^n}{n^4\pi^4}\text{ for }n\ge1\tag{2} $$ The contour we will use goes from $+i\infty$ to $0$ then to $+\infty$ with infinitesimal clockwise semi-circles to avoid the real singularities. By contour integration, the principal value of integral along the real axis from $0$ to $\infty$ is equal to the integral along the imaginary axis from $0$ to $i\infty$ plus $\pi i$ times the sum of the residues at $x_n$ for $n\ge1$ plus $2\pi i$ times the residue at $x_0$. The sum of the residues obviously converges.

Combining the integral along the imaginary axis, the residues along the real axis (which only combine an imaginary part), and the residue at $x_0$, we get the Cauchy Principal Value for the integral: $$ \begin{align} \mathrm{PV}\int_0^\infty\frac{\mathrm{d}x}{1+x^4\sin(x)} &=\int_0^\infty\frac{i\,\mathrm{d}x}{1+ix^4\sinh(x)}\\ &+\pi i\sum_{n=1}^\infty\frac1{4x_n^3\sin(x_n)+x_n^4\cos(x_n)}\\ &+2\pi i\frac1{4x_0^3\sin(x_0)+x_0^4\cos(x_0)}\\ &=\int_0^\infty\frac{x^4\sinh(x)\,\mathrm{d}x}{1+x^8\sinh^2(x)}\\[9pt] &+0.85233885641320594757\tag{3} \end{align} $$ Thus, although the integrand is not absolutely integrable, it does have a convergent Cauchy Principal Value. Using $(3)$, Mathematica 8 gives the numerical approximation $$ \mathrm{PV}\int_0^\infty\frac{\mathrm{d}x}{1+x^4\sin(x)}=1.14619893142224184361\tag{4} $$