What is the Limit of the Integral in the Small x Regime for y approaching 1?

[blue2script]

Dear all,
I want to calculate the following integral

$$\int_{-\infty}^0 dk \frac{k\left(\frac{k^2-m^2}{k}\cos\frac{2(x M - k)c_0}{m y} + m\sin\frac{2(x M - k)c_0}{m y} + \frac{k^2+m^2}{2k}\right)}{\sinh^2\frac{(x M - k)\pi}{2my}((k^2 - m^2)^2 + 4 k^2 m^2 y^2)}$$

in the limit $$y\to 1$$ to examine the small x regime (x > 0, x << 1). However, $$c_0$$ is given by

$$c_0 = \frac{1}{2}\operatorname{arctanh}y$$

so it diverges in the limit $$y\to 1$$. But then I would state that we may neglect the cosine and sine terms since they oscillate so rapidly that there contribution to the integral vanishes. My professor however, with whom I discussed this matter, says I am not eligible to do that since I want to examine the low-x regime where I get a pole in the limit $$x\to 0$$. Than my arguing would not be true.

I told him I would try to give this one a rigourious mathematical treatment. But then 1) I can't see what is wrong with my arguments since I am not examing x = 0 but only small but non-zero x where there is no pole and 2) it looks so obvious to me that I don't really know how to treat this on solid mathematical grounds.

Thats why I would really appreciate a discussion about this integral in the limit $$y\to 1$$. Hopefully some of you has some idea how to treat this.

A big thanks in advance!
Blue2script

 

[flatmaster]

Yowzers, quite an integral.

If you're assuming small x...

Mx-k == -kAs long as Mx <<< k

 

[Mute]

I don't know if it would help any, but you could combine the cosine and sine using the identity

$$A\cos\phi + B\sin\phi = \sqrt{A^2 + B^2}\sin\left(\phi + \tan^{-1}\left(\frac{A}{B}\right)\right)$$

(note: if A/B < 0, then a phase factor of pi must be added)