Question

(a) The improper integrals \(\int_{1}^{\infty} \frac{1}{x} d x \quad\) and \(\int_{1}^{\infty} \frac{1}{x^{2}} d x\) diverge and converge, respectively. Describe the essential differences between the integrands that cause one integral to converge and the other to diverge. (b) Sketch a graph of the function \(y=\sin x / x\) over the interval \((1, \infty)\). Use your knowledge of the definite integral to make an inference as to whether or not the integral \(\int_{1}^{\infty} \frac{\sin x}{x} d x\) converges. Give reasons for your answer. (c) Use one iteration of integration by parts on the integral in part (b) to determine its divergence or convergence.

Short answer

First integral \(\int_{1}^\infty \frac{1}{x} dx\) diverges. Second integral \(\int_{1}^\infty \frac{1}{x^{2}} dx\) converges. The function \(y=\sin x / x\) over the interval \((1, \infty)\) diminishes as \(x\) gets larger, predicting that the integral \(\int_{1}^\infty \frac{\sin x}{x} dx\) also converges. Integration by parts confirmed this intuition, as the integration resulted in the sum of a definite integral from 1 to ∞ of a function smaller than \(\frac{1}{x^2}\) which we know converges, and thus the original integral also converges.

Step-by-step solution

Evaluate First Two Integrals

Evaluate the improper integrals\(\int_{1}^{\infty} \frac{1}{x} d x \) and \(\int_{1}^{\infty} \frac{1}{x^{2}} d x\). Using standard integration, the first integral evaluates to \([ \ln |x| ]_{1}^{\infty}\) and after evaluating the limits results in \( \infty - \ln 1\) breaking the limit rule, i.e., infinity minus a finite number, hence, it diverges. The second integral evaluates to \([- \frac{1}{x} ]_{1}^{\infty}\) and after evaluating the limits it is \( -0 - -1 = 1 \) so it converges.

Analyze Integrands

The primary difference between the integrands that is causing one integral to diverge, and the other to converge is the exponent of \(x\) in the denominator. As the exponent increases, the value of the denominator increases significantly causing the whole fraction to approach 0 faster, therefore causing the integral to converge.

Sketch Function and Analyze

Sketch the function \(y = \sin x / x\) over the interval \((1, \infty)\) and infer the convergence or divergence of \(\int_{1}^{\infty} \frac{\sin x}{x} d x\). For \(x > 1\), \(\sin x / x\) oscillates between positive and negative values. As \(x\) gets larger, the amplitude of oscillations diminishes due to the division by \(x\), suggesting the integral of the function could converge.

Integrate By Parts

To prove the inference in Step 3, we can use integration by parts on the integral \(\int_{1}^{\infty} \frac{\sin x}{x} d x\). The integration by parts formula is \(\int udv = uv - \int vdu\). Let \(u=\frac{1}{x}\) and \(dv=\sin x\ dx\). Then \(du=-\frac{1}{x^2}dx\) and \(v=-\cos x\). Using these values in the integration by parts formula, we get \( \left[-\frac{\cos x}{x}\right]_1^\infty + \int_1^\infty \frac{\cos x}{x^2} dx \). The first term evaluates to 0 because both terms within the brackets tend to 0. The second term is an integral of a function smaller than \(\frac{1}{x^2}\) over [1, ∞) which, as we know from before, converges. Therefore, the integral in question also converges.