It is possible to obtain a more direct proof using other idea due to Professor Ebert.

It is well known that

$\hat{A} = \prod _ i{\frac {x_{{i}}/2}{\sinh \left( x_{{i}}/2 \right) }}= 1+\hat{A}_{1}+\hat{A}_{2}+\hat{A}_{3}+....$

where

$\hat{A}_{1}=-\frac{1}{24}p_{{1}}$

$\hat{A}_{2} = -{\frac {1}{1440}}\,p_{{2}}+{\frac {7}{5760}}\,{p_{{1}}}^{2}$

$\hat{A}_{3} = {\frac {11}{241920}}\,p_{{1}}p_{{2}}-{\frac {1}{60480}}\,p_{{3}}-{
\frac {31}{967680}}\,{p_{{1}}}^{3} $

From other side

$ {\frac {\hat{A}}{\hat{M}}}= \prod _ i{\frac {x_{{i}}}{\sinh \left( x_{{i}} \right) }} = 1+ [{\frac {\hat{A}}{\hat{M}}}]_{1}+ [{\frac {\hat{A}}{\hat{M}}}]_{2}+[{\frac {\hat{A}}{\hat{M}}}]_{3}+....$

where

$[{\frac {\hat{A}}{\hat{M}}}]_{1} = -\frac{1}{6}p_{{1}}$

$[{\frac {\hat{A}}{\hat{M}}}]_{2} = {\frac {7}{360}}\,{p_{{1}}}^{2}-{\frac {1}{90}}\,p_{{2}} $

$[{\frac {\hat{A}}{\hat{M}}}]_{3} = {\frac {11}{3780}}\,p_{{1}}p_{{2}}-{\frac {31}{15120}}\,{p_{{1}}}^{3}-
{\frac {1}{945}}\,p_{{3}}
$

Then we have

$[{\frac {\hat{A}}{\hat{M}}}]_{1} = 4\hat{A}_{1}$

$[{\frac {\hat{A}}{\hat{M}}}]_{2} = 16\hat{A}_{2}$

$[{\frac {\hat{A}}{\hat{M}}}]_{3} = 64\hat{A}_{3}$

$[{\frac {\hat{A}}{\hat{M}}}]_{n} = 4^{n}\hat{A}_{n}$

From the last equation we obtain that given that

$[\hat{A}(\mathbb HP^m)]_{4m} = 0$

then

$[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }]_{4m} = 0$.

Besides of this, the paragraph

extracted from Baer, page 30; can be simplified to the integrality of

$2^{l+\frac{n}{2}}\int_{X}\hat{A}_{\frac{n}{4}}(TX)$.

As an application of the last result it is possible to prove that $CP^4$ cannot be immersed in $R^{11}$. Computing the last integral with $l=1$ and $n=8$ we obtain

$2^{5}\int_{CP^4}\hat{A}_{2}(TCP^4)= \frac{3}{4} $

which is not an integer.

In general for $CP^4$ the Mayer integral takes the form

$2^{l+4}\int_{CP^4}\hat{A}_{2}(TCP^4) = 3({2})^{l-3}$

Then this last integral indicates that $CP^4$ cannot be immersed in $R^{10}, R^{11}, R^{12}, R^{13}$. For $l\geq 3$ there is no topological obstruction to the existence of immersion of $CP^4$ in $R^{8+2l+1}$.

More in general, in the case of $CP^{2m}$ the Mayer integral is evaluated as

$2^{l+2m}\int_{CP^{2m}}\hat{A}_{m}(TCP^{2m}) = (-1)^m2^{l-2m}{2\,m\choose m}$.