How can this converge?!

Calculus Level 3

Consider the piecewise function $f(x)$ defined as follows:

$f(x) = \begin{cases} \dfrac{-\cos x}{x} +\sin(1) +\cos(1) &\quad 0\leq x <1 \\ \phantom0 \\ \ \ \ \sin \dfrac{1}{x} &\quad x\geq1 \end{cases}$

If the Cauchy principal value of $\displaystyle \int_{0}^{\infty} f(x) \, dx$ is equal to $A+ \cos(B)$ for positive integers $A$ and $B$ , submit your answer as $A+B.$

The answer is 2.

1 solution

Amal Hari
Dec 17, 2020

Consider the integral $\displaystyle \int_{1}^{\infty} \sin\frac{1}{x} dx \\$ Substituting $\frac{1}{x} =u$ we have $\displaystyle \int_{1}^{\infty} \sin\frac{1}{x} dx = \int_{1}^{0} \frac{-\sin u}{u^{2}} du = \frac{\sin u}{u}\Biggr|_{1}^{0} -\int_{1}^{0} \frac{\cos u }{u} = 1- \sin(1 ) + \int_{0}^{1} \frac{\cos u }{u}$

$\displaystyle \int_{1}^{\infty} \sin\frac{1}{x} dx = 1- \sin(1 ) + \int_{0}^{1} \frac{\cos u }{u} du\\$ $\displaystyle \int_{1}^{\infty} \sin\frac{1}{x} dx - \int_{0}^{1} \frac{\cos u }{u} du = 1- \sin(1 ) \\$

Now consider the integral in the problem:

$\displaystyle \int_{0}^{\infty} f(x) dx =\displaystyle - \int_{0}^{1} \frac{\cos x }{x} dx +\int_{0}^{1} \left(\cos(1) +\sin(1) \right)dx+ \int_{1}^{\infty} \sin\frac{1}{x} dx \\$ With regards to our result from the intro we can rewrite this as $\displaystyle \int_{0}^{\infty} f(x) dx =\displaystyle 1- \sin(1 ) +\int_{0}^{1} \left(\cos(1) +\sin(1) \right)dx =1+\cos 1 \\$

$A= 1, B=1, A+B=2$

How can this converge?!

The answer is 2.

1 solution

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