# Limit of an integral

1. Dec 19, 2006

### traianus

Hello,
Suppose to have the following integral:

$$\int \limits _{-H/2}^{+H/2}f(z) \frac{H-2z}{\left[\left(b - y\right)^2 + \left(H/2 - z\right)^2\right]^2}dz$$

Suppose that $$f(z)$$ does NOT have a crazy behavior and that does not go to infinity anywhere and that it is continuos. I do not know a priori the expression of $$f(z)$$.

Now the question: what is the limit of the integral when the parameter $$H$$ (which appears in the limits and integrand) goes to $$+\infty$$ ???

2. Dec 20, 2006

### quasar987

In my opinion, it is not possible to say anything without knowing explicitely f(z).

3. Dec 20, 2006

### traianus

That is the issue I have. Let's assume the function to be odd or even or neither.

4. Dec 20, 2006

### StatusX

Try integrating by parts once (assuming f is differentiable).

5. Dec 20, 2006

### traianus

If I integrate by parts, what do I gain? I do not see it????

6. Dec 20, 2006

### StatusX

You should get the derivative of f time a spike centered at H/2. Pay attention to the boundary terms as well. You can get a nice result if you assume f' is continuous and vanishes at infinity.

Last edited: Dec 20, 2006
7. Dec 20, 2006

### tehno

Nothing concretely can be concluded about convergence or divergence without specifying f(z) I affraid.
Suggestion however:
You might want to try some classical range tests with f(z) being constant ,logarithm and exponential funtions to get the feeling of behaviour.

8. Dec 20, 2006

### traianus

Could you please StatusX post your procedure here? I have the feeling that tehno is right and f(z) must be specified. But we can also ask (this can be useful too) what are the properties of the function f(z) to NOT have infinite limit.

9. Dec 20, 2006

### StatusX

All I can really say is that if f' vanishes at infinity, you can show the integral you get after integrating by parts vanishes in the limit. Then you are left with the boundary terms, which should be easy, as long as f(z) asympototes to some limit as z->infinity (which, incidentally, implies the first assumption).

10. Dec 20, 2006

### traianus

If I unserstood correctly, when the derivative goes to zero the we have only the boundary term; however, the boundary term goes to zero too because it has H at the denominator. So under the assumptions that StatusX made the limit is zero.

11. Dec 20, 2006

### StatusX

No, check the boundary term again, the H should drop out.

12. Dec 22, 2006

### balakrishnan_v

If f converges to a point say $$f_{\infty}$$ then the answer would be
$$\frac{f_{\infty}}{(b-y)^2}$$

13. Dec 27, 2006