integral of sin(x)/x as x approaches infinity,?

Question

but not like on this web page http://mathforum.org/library/drmath/view/67312.html
thx

Answer 1

This can be done by means of Laplace Transform. Since the Laplace transform of the function sin(x) is the function 1/(1 + s^2), it follows from the properties of laplace Transform that

Int (o to oo) sin(x)/x dx = Integral (o to oo) ds/(1 + s^2) = arctan (s) [0 to oo] = pi/2 - 0 = pi/2

Answer 2

integral of sin(x)/x as x approaches infinity = pi/2

integral of sin(x)/x from 0 to x = Si(x)
when x approaches infinity then Si(x) = pi/2

Answer 3

Firstly as sin(x) is bounded by [-1,1] we have
| sin(x)/x | < 1/|x| (where< is less than or equal)
lim(x,inf) |sin(x)/x| < lim(x,inf) 1/x = 0
=> lim(x,inf) (sin(x)/x) = 0
So there will be no contribution from the integral in the limit as x approches inf. To solve the actual integreation from some positive constant k to inf you will have to power series and piecewise integration

int(k,inf) sin(x)/x dx = int(k,inf) sum(n=0,inf) [(-1)^n*x^2n]/(2n+1)!
= sum(n=0,inf) [ (-1)^(n+1) * k^(2n+1)]/((2n+1)*(2n+1)!)

Answer 4

This limit does not exist.

Sin(x) does not coverge as x -> infinity

Using x as a denominator would tell us that the entire term (sin (x) / x ) -> 0 as x -> infinity

thus.

Answer 5

Donald Trump's bank account.

Answer 6

what do u mean by integral?

Answer 7

It doesn't exist.

Answer 8

i dont know if you can do that?