Calculate
If you come to the conclusion that the integral diverges, answer with .
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We will focus us on the integral I = ∫ 0 ∞ 1 − x sin ( x 1 ) d x . Substitute x = u 1 , then d x = − u 2 1 d u and so we notice that I = − ∫ ∞ 0 u 3 u − sin ( u ) d u = ∫ 0 ∞ u 3 u − sin ( u ) d u . We will now use integration by parts. We will differentiate the part u − sin ( u ) and integrate the part u 3 1 to get I = − 2 u 2 u − sin ( u ) ∣ ∣ ∣ ∣ 0 ∞ + ∫ 0 ∞ 2 u 2 1 − cos ( u ) d u . We will repeat the last step with the new integral to get I = − 2 u 2 u − sin ( u ) − 2 u 1 − cos ( u ) ∣ ∣ ∣ ∣ 0 ∞ + 2 1 ⋅ ∫ 0 ∞ u sin ( u ) d u . We will show that 2 u 2 u − sin ( u ) + 2 u 1 − cos ( u ) ∣ ∣ ∣ ∣ 0 ∞ = 0 . It is not difficult to see that lim u → ∞ [ 2 u 2 u − sin ( u ) + 2 u 1 − cos ( u ) ] = 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 u → 0 lim 2 u 2 u − sin ( u ) = u → 0 lim 4 u 1 − cos ( u ) = u → 0 lim 4 sin ( u ) = 0 , which shows us that indeed lim u → 0 [ 2 u 2 u − sin ( u ) + 2 u 1 − cos ( u ) ] = 0 and so 2 u 2 u − sin ( u ) + 2 u 1 − cos ( u ) ∣ ∣ ∣ ∣ 0 ∞ = 0 . We will now evaluate the integral J = ∫ 0 ∞ u sin ( u ) d u . We will do this by using Feynman's trick: differentiation under the integral sign. Define J ( t ) = ∫ 0 ∞ u e t u ⋅ sin ( u ) d u for t ≤ 0 . Then d t d J ( t ) = d t d ∫ 0 ∞ u e t u ⋅ sin ( u ) d u = ∫ 0 ∞ ∂ t ∂ u e t u ⋅ sin ( u ) d u = ∫ 0 ∞ e t u ⋅ sin ( u ) d u . By again using integration by parts, we can show that J ′ ( t ) = 1 − t 2 ⋅ J ′ ( t ) , from which follows that J ′ ( t ) = 1 + t 2 1 . We conclude that J ( t ) = ∫ 1 + t 2 d t = arctan ( t ) + C . As for all x ∈ ( 0 , ∞ ) , ∣ ∣ ∣ ∣ x sin ( x ) ∣ ∣ ∣ ∣ ≤ 1 , it follows that 0 ≤ t → − ∞ lim ∣ J ( t ) ∣ = t → − ∞ lim ∣ ∣ ∣ ∣ ∫ 0 ∞ u e t u ⋅ sin ( u ) d u ∣ ∣ ∣ ∣ ≤ t → − ∞ lim ∫ 0 ∞ ∣ ∣ ∣ ∣ u e t u ⋅ sin ( u ) ∣ ∣ ∣ ∣ d u ≤ t → − ∞ lim ∫ 0 ∞ e t u d u = t → − ∞ lim t e t u ∣ ∣ ∣ ∣ u = 0 u = ∞ = t → − ∞ lim t − 1 = 0 , so lim t → − ∞ J ( t ) = lim t → − ∞ arctan ( t ) + C = 0 , which implies that C = 2 π . This means that J = J ( 0 ) = arctan ( 0 ) + 2 π = 2 π .
By using all the information we got before, we can conclude that π 1 ⋅ ∫ 0 ∞ [ 1 − x sin ( x 1 ) ] d x = π 1 ⋅ 2 1 ⋅ 2 π = 4 1 .