An integral containing a well known limit.

Calculus Level 5

Calculate

1 π 0 [ 1 x sin ( 1 x ) ] d x . \frac{1}{\pi} \int_0^{\infty} \left[1-x\sin\left(\frac{1}{x}\right)\right]\mathrm{d}x.

If you come to the conclusion that the integral diverges, answer with 2019 2019 .


The answer is 0.25.

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1 solution

Joël Ganesh
Aug 13, 2019

We will focus us on the integral I = 0 1 x sin ( 1 x ) d x . I = \int_0^\infty 1-x\sin\left(\frac{1}{x}\right)\mathrm{d}x. Substitute x = 1 u x = \frac{1}{u} , then d x = 1 u 2 d u \mathrm{d}x = -\frac{1}{u^2}\mathrm{d}u and so we notice that I = 0 u sin ( u ) u 3 d u = 0 u sin ( u ) u 3 d u . I = -\int_\infty^0 \frac{u-\sin(u)}{u^3} \mathrm{d}u = \int_0^\infty \frac{u-\sin(u)}{u^3} \mathrm{d}u. We will now use integration by parts. We will differentiate the part u sin ( u ) u-\sin(u) and integrate the part 1 u 3 \frac{1}{u^3} to get I = u sin ( u ) 2 u 2 0 + 0 1 cos ( u ) 2 u 2 d u . I = \left. -\frac{u-\sin(u)}{2u^2} \right|_0^\infty + \int_0^\infty \frac{1-\cos(u)}{2u^2} \mathrm{d}u. We will repeat the last step with the new integral to get I = u sin ( u ) 2 u 2 1 cos ( u ) 2 u 0 + 1 2 0 sin ( u ) u d u . I = \left. -\frac{u-\sin(u)}{2u^2} - \frac{1-\cos(u)}{2u} \right|_0^\infty + \frac{1}{2} \cdot \int_0^\infty \frac{\sin(u)}{u} \mathrm{d}u. We will show that u sin ( u ) 2 u 2 + 1 cos ( u ) 2 u 0 = 0. \left. \frac{u-\sin(u)}{2u^2} + \frac{1-\cos(u)}{2u} \right|_0^\infty = 0. It is not difficult to see that lim u [ u sin ( u ) 2 u 2 + 1 cos ( u ) 2 u ] = 0 \lim_{u\to\infty} \left[ \frac{u-\sin(u)}{2u^2} + \frac{1-\cos(u)}{2u}\right] = 0 . This can be calculated with the squeeze theorem. We will focus us on the other required limit to evaluate. With l'Hôpital's rule we find that lim u 0 u sin ( u ) 2 u 2 = lim u 0 1 cos ( u ) 4 u = lim u 0 sin ( u ) 4 = 0 , \lim_{u\to 0} \frac{u-\sin(u)}{2u^2} = \lim_{u\to 0} \frac{1-\cos(u)}{4u} = \lim_{u\to 0} \frac{\sin(u)}{4} = 0, which shows us that indeed lim u 0 [ u sin ( u ) 2 u 2 + 1 cos ( u ) 2 u ] = 0 \lim_{u\to 0} \left[ \frac{u-\sin(u)}{2u^2} + \frac{1-\cos(u)}{2u}\right] = 0 and so u sin ( u ) 2 u 2 + 1 cos ( u ) 2 u 0 = 0. \left. \frac{u-\sin(u)}{2u^2} + \frac{1-\cos(u)}{2u} \right|_0^\infty = 0. We will now evaluate the integral J = 0 sin ( u ) u d u . J = \int_0^\infty \frac{\sin(u)}{u} \mathrm{d}u. We will do this by using Feynman's trick: differentiation under the integral sign. Define J ( t ) = 0 e t u sin ( u ) u d u J(t) = \int_0^\infty \frac{e^{tu} \cdot \sin(u)}{u} \mathrm{d}u for t 0 t \leq 0 . Then d d t J ( t ) = d d t 0 e t u sin ( u ) u d u = 0 t e t u sin ( u ) u d u = 0 e t u sin ( u ) d u . \frac{\mathrm{d}}{\mathrm{d}t} J(t) = \frac{\mathrm{d}}{\mathrm{d}t} \int_0^\infty \frac{e^{tu} \cdot \sin(u)}{u} \mathrm{d}u = \int_0^\infty \frac{\partial}{\partial t} \frac{e^{tu} \cdot \sin(u)}{u} \mathrm{d}u = \int_0^\infty e^{tu} \cdot \sin(u) \mathrm{d}u. By again using integration by parts, we can show that J ( t ) = 1 t 2 J ( t ) J'(t) = 1 - t^2 \cdot J'(t) , from which follows that J ( t ) = 1 1 + t 2 J'(t) = \frac{1}{1+t^2} . We conclude that J ( t ) = d t 1 + t 2 = arctan ( t ) + C . J(t) = \int \frac{\mathrm{d} t}{1+t^2} = \arctan(t) + C. As for all x ( 0 , ) x \in (0,\infty) , sin ( x ) x 1 , \left|\frac{\sin(x)}{x}\right| \leq 1, it follows that 0 lim t J ( t ) = lim t 0 e t u sin ( u ) u d u lim t 0 e t u sin ( u ) u d u lim t 0 e t u d u = lim t e t u t u = 0 u = = lim t 1 t = 0 , 0 \leq \lim_{t\to -\infty} |J(t)| = \lim_{t\to -\infty} \left|\int_0^{\infty} \frac{e^{tu} \cdot \sin(u)}{u}\mathrm{d}u\right| \leq \lim_{t\to -\infty} \int_0^{\infty} \left|\frac{e^{tu} \cdot \sin(u)}{u}\right|\mathrm{d}u \leq \lim_{t\to -\infty} \int_0^{\infty} e^{tu} \mathrm{d}u = \lim_{t\to -\infty} \left.\frac{e^{tu}}{t}\right|_{u=0}^{u=\infty} = \lim_{t\to -\infty} \frac{-1}{t} =0, so lim t J ( t ) = lim t arctan ( t ) + C = 0 \lim_{t\to -\infty} J(t) = \lim_{t\to -\infty} \arctan(t) + C = 0 , which implies that C = π 2 C = \frac{\pi}{2} . This means that J = J ( 0 ) = arctan ( 0 ) + π 2 = π 2 J = J(0) = \arctan(0) + \frac{\pi}{2} = \frac{\pi}{2} .

By using all the information we got before, we can conclude that 1 π 0 [ 1 x sin ( 1 x ) ] d x = 1 π 1 2 π 2 = 1 4 . \frac{1}{\pi} \cdot \int_0^\infty \left[1-x\sin\left(\frac{1}{x}\right)\right]\mathrm{d}x = \frac{1}{\pi} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \boxed{\frac{1}{4}}.

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