Question

what do you say about the integral of sin(1/x)?

Answer 1

Write it as a summation:
\[\int\limits \left[ \sum_{k=0}^{\infty}\frac{ (-1)^{k} \left( \frac{1}{x} \right) ^{2k+1}}{{(2k+1)!}} \right]dx\]

Answer 2

Use the alternating series estimatino theorem to estimate the value.

Answer 3

integral sin(1/x) dx = x sin(1/x)-cos(1/x)+constant

Alternate Form:

-Ci(1/x)+1/2 i e^(-i/x) x-1/2 i e^(i/x) x+constant

Series expansion of the integral at x=0:

-2 i pi floor((arg(x)+pi)/(2 pi))+i pi floor((arg(x))/pi+1/2)+O(x^9)+(x+O(x^7)) sin(1/x)+(-x+2 x^3-24 x^5+O(x^7)) sin(1/x)+(x^2-6 x^4+120 x^6+O(x^7)) cos(1/x)

Series expansion of the integral at x=∞:

(-log(1/x)-gamma+1)+1/(12 x^2)-1/(480 x^4)+1/(30240 x^6)-1/(2903040 x^8)+1/(399168000 x^10)+O((1/x)^11)

I hope this answers your question!

Answer 4

Its not -cos(1/x) its minus the cosine integral. Which is an unsolvable integral. Which must be approximated.

Answer 5

It is not possible to find the exact value you can approximate the value by putting it inbetween two known integrals....

Answer 6

By using the alternating series estimation theorem ________.________