Integral xsinx limits 0 to infinity

In summary, the conversation discusses attempts to evaluate an integral involving a sinusoidal function and its implications in the context of x-ray scattering. The speaker has tried various methods, including integration by parts and using Euler's formula, but has been unable to get the integral to converge. They also mention that this issue may be related to previous calculations.
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Homework Statement



So I'm trying to evaluate the following integral:

[tex] 4\pi r^2{\int_0}^\infty r^2\frac{\sin{sr}}{sr}dr[/tex]

which after canceling out one of the r's, gives an integral similar to that of xsinx.
I need to show that this integral vanishes for all values of s that are not 0. In other words, this integral has a non-zero value only when s is equal to zero. This leads me to think that this integral is some sort of a delta function.

By integration by parts I get the following:

[tex] \frac{4\pi}{s^3}(\sin{sr} -sr\cos{sr}){\mid_0}^\infty [/tex]

I've tried to play around with trig identities when evaluating the limits of integration, but can't seem to get anything that doesn't involve trig functions to get it to converge, since the limit of cos(x) or sin(x) as x goes to infinity is undefined due to it's oscillatory nature. I have also tried evaluating this integral by first using euler's formula to convert the sin function into powers of e, but this results in similar problems when evaluating the limits. Any suggestions would be appreciated. This problem comes from x-ray scattering, so maybe someone familiar with this subject has seen this.

Homework Equations





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  • #2
Your antiderivative looks correct. It appears that the integral doesn't converge though. No amount of rewriting (via trig identities, algebra, etc.) is going to force an integral that doesn't converge to converge.

EDIT: It's been awhile since I've taken any physics courses, but I remember in E&M always coming across integrals that blew up. That was usually a sign I screwed up earlier in my calculations.
 
Last edited:

Related to Integral xsinx limits 0 to infinity

What is the integral of xsinx from 0 to infinity?

The integral of xsinx from 0 to infinity is equal to 1/2, or 0.5.

Why is the integral of xsinx from 0 to infinity equal to 1/2?

The integral of xsinx from 0 to infinity is equal to 1/2 because the function xsinx oscillates between positive and negative values, resulting in an average value of 1/2 over the entire interval.

How do you solve the integral of xsinx from 0 to infinity?

To solve the integral of xsinx from 0 to infinity, you can use the integration by parts method, where u=x and dv/dx=sinx. By applying this method, you will arrive at the result of 1/2.

What is the significance of the limits being from 0 to infinity in the integral of xsinx?

The limits of 0 to infinity in the integral of xsinx represent the entire domain of the function. This allows for a more accurate calculation of the average value of the function over the entire interval.

Can the integral of xsinx from 0 to infinity be approximated using numerical methods?

Yes, the integral of xsinx from 0 to infinity can be approximated using numerical methods such as the trapezoidal rule or Simpson's rule. These methods divide the interval into smaller sections and approximate the integral using the values of the function at each section.

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