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

# 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:

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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:

- (\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} ]

- (\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).

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