Is There a Method for Integrating Sinx/x Between 0 and Infinity?

[heidernetk]

Can you help me by giving me a method or solution



integrating sinx/x between (0,infinty)

please help me

 

[pixel01]

sinx/x does not have the elementary primitive function. You can calculate it numerically by a program. I used Matlab and calculated it as :1.571

 

[morphism]

sinx/x does not have the elementary primitive function. You can calculate it numerically by a program. I used Matlab and calculated it as :1.571

That's an approximation to $\pi / 2$. Actually, the result
$$\int_0^\infty \frac{\sin x}{x} \: dx = \frac{\pi}{2}$$
is somewhat well-known. It can be proved using complex analysis or Fourier analysis. heidernetk, do you have experience with either?

(I also seem to recall seeing another way of doing this that doesn't use complex/Fourier analysis, but I can't remember how the argument went.)

 

[robert Ihnot]

Using contour integration you can solve this. The trick is to use

$$\int\frac{e^{iz}}{z} \: dz$$ and find the imaginary part. This can be done using residue theory. However, I took that course long ago and don't know all the steps.

 

[JonF]

If i recall you can take the integral over a quarter circle with a radius of R in the complex plane and Quadrant I. Then look at the over all path compared to the path partitioned into linear components an the circular component as R tends to infinity.

 

[AlphaNumeric]

sinx/x does not have the elementary primitive function. You can calculate it numerically by a program. I used Matlab and calculated it as :1.571

There wouldn't be much point in asking the question for a maths homework/question sheet if the answer was "Done by computer". For integrals such as sinx/x, particularly when the limits involve infinity, complex analysis is often the way to go.

However, I took that course long ago and don't know all the steps.

For the benefit of the original poster mostly :

1. Make the contour the semicircle through the upper half plane but with a slight bump around the origin due to singularity (due to cos part of e^iz)
2. Split the contour into 4 parts, from epsilon to R along the Real axis, the semicircle arc of radius R round to -R on the Real axis, from -R to -epsilon and then a semicircle from -epsilon to +epsilon. Doesn't matter which side of the origin that little contour goes, the orientated nature of the path sorts out the signs and residue contributions.
3. Use the residue theorem to relate this contour integral to the sum of residues within the contour.
4. Take R->infinity and use Jordan's Lemma to justify it giving zero contribution. By the fact the imaginary part of e^iz/z is even, you end up having the integrals from -infinity to -epsilon being equal to +epsilon to +infinity.
5. Thus means you just have to compute the little contour and the residue at zero. Stick it in the equation due to the Residue Theorem, rearrange and all going well and a following wind, you should get the answer.

On the scale of things, one of the much nicer contour integrals one ever does. I used to freakin' hate ones involving branch cuts from say -1 to +1.

 

[minase]

Is there any way to solve this using integration by parts.

 

[Gib Z]

No there isn't and this is really pointless but the anti derivative is defined to be the non elementary Si(x) function.