What is #int_1^oo 1/x^3-sinx/x^3 dx#?

Question

What is #int_1^oo 1/x^3-sinx/x^3   dx#?

Paisley Johnson · Accepted Answer

# int_0^oo 1/x^3 - (sinx)/x^3 dx = 0.1215# (4dp) graph{1/x^3 - (sinx)/x^3 [-3, 8, -2, 2]} From the graph it would appear that the improper integral does converge. However the integrand does not have an elementary anti-derivative. Numerical methods can be used to evaluate the integral which has the value: # int_0^oo 1/x^3 - (sinx)/x^3 dx = 0.12146998 ...#

Isabella Ackerman · Answer

undefined To find the integral $\int_{1}^{\infty} \frac{1/x^3 - \sin x}{x^3} \, dx$, we first need to integrate each term separately.

Let's break down the integral:

$$
\int_{1}^{\infty} \frac{1/x^3 - \sin x}{x^3} \, dx = \int_{1}^{\infty} \frac{1}{x^3} \, dx - \int_{1}^{\infty} \frac{\sin x}{x^3} \, dx
$$

Now, let's integrate each term:

1. $\int_{1}^{\infty} \frac{1}{x^3} \, dx$

Using the power rule for integration, we get:

$$
\int_{1}^{\infty} \frac{1}{x^3} \, dx = \left[ -\frac{1}{2x^2} \right]_{1}^{\infty}
$$

As $x$ approaches infinity, the expression tends to zero, and at $x = 1$, it is $-\frac{1}{2}$. Thus, the integral becomes:

$$
\left(0 - \left(-\frac{1}{2}\right)\right) = \frac{1}{2}
$$

2. $\int_{1}^{\infty} \frac{\sin x}{x^3} \, dx$

This integral requires more advanced techniques, specifically integration by parts. Let $u = \frac{1}{x^3}$ and $dv = \sin x \, dx$.

Then, $du = -\frac{3}{x^4} \, dx$ and $v = -\cos x$.

Applying the integration by parts formula:

$$
\int udv = uv - \int vdu
$$

We get:

$$
\int_{1}^{\infty} \frac{\sin x}{x^3} \, dx = \left[ -\frac{\cos x}{x^3} \right]_{1}^{\infty} - \int_{1}^{\infty} \frac{3\cos x}{x^4} \, dx
$$

As $x$ approaches infinity, $\frac{\cos x}{x^3}$ approaches zero, and at $x = 1$, it is $-\frac{\cos 1}{1^3}$. The remaining integral requires further calculations.

Thus, the integral $\int_{1}^{\infty} \frac{1/x^3 - \sin x}{x^3} \, dx$ evaluates to $\frac{1}{2} + \frac{\cos 1}{1^3} - \int_{1}^{\infty} \frac{3\cos x}{x^4} \, dx$.