can some one integrate (sin x)/x for me?

Question

limits from e to 3pi

Answer 1

As stated above, there is not a "closed form" solution for \int sin(x)/x dx. Typically, we approximate the value with series. In a calculus book, you'd find a probem like this in the section where they discussion differentiation and integration of power series. Power series have the property that they can be differentiated and integrated term-by-term on their interval of convergence.

First, sin(x)=\sum_{k=0)^{infinity} (-1)^k x^(2k+1)/(2k+1)! for all x.

Then,
sin(x)/x=\sum_{k=0)^{infinity} (-1)^k x^(2k)/(2k+1)!

Hence,
\int_{0}^{x} sin(t)/t dt=
\int_{0}^{x} (\sum_{k=0)^{infinity} (-1)^k t^(2k)/(2k+1)!) dt=
\sum_{k=0}^{infinity} (-1)^k x^(2k+1)/[(2k+1)(2k+1)!]

Answer 2

Sin(x)/x has no elementary integral. In other words it cannot be integrated using standard methods.

One way of doing it would be to use a numerical method, but this will only give the definite integral because with numerical methods you use limits.

Answer 3

That is impossible to express as a regular integral, but i'm pretty sure that there's a way that you can represent the answer using power series. I'm guessing that if you're getting to sinx/x, you're probably in entry-level calculus, and wouldn't be expected to do that. If you are doing the Sandwich Theorem (also called the Squeeze Theorem), then I would say to you that for your purposes, this is non-integrable.

Answer 4

It is one of the
curious facts of calculus that the antiderivatives of many simplelooking
functions cannot be expressed in terms of the “elementary
functions,” that is, polynomials and ratios of polynomials,
exponential and trigonometric functions and their inverses, and
any finite combination of these functions. The function êsin xë=x
belongs to this group, as do êcos xë=x, ex=x, and ex2 . This, of
course, does not mean that the antiderivatives of these functions
do not exist—it only means that they cannot be expressed
in “closed form” in terms of the elementary functions. Indeed,
the above integral, regarded as a function of its upper limit x,
defines a new, “higher” function known as the sine integral and
denoted by Si êxë:
http://www.pupress.princeton.edu/books/maor/chapter_10.pdf

Answer 5

i do not think the indefinite integral can be easily evaluated.

But the definite integral with limits 0 to infinity is evaluated to pi/2

Usually complex residue integration is used
or the result is deduced from laplace/fourier transform

Answer 6

What are you going over in your calculus class? Are you just applying the quotient rule? Are you evaluating indefinite integrals? What do you need?

Answer 7

Try integration by parts.

Answer 8

If you can integrate SIN1. The X cancel eachother out.

Answer 9

i think u r in 12th.Sorry man, there is a certain level to which u can answer Qs . in the university u will be able 2 do it as higher level of maths is taught.

Answer 10

For problems of this type I recommend
http://www.quickmath.com/
Sadly in this case you are out of luck but give the site a play, it is incredible what it can do.