Consider a canonical transformation from $(q,p)$ to $(Q,P)$ depending upon a continuous parameter $\alpha$ such that: $$Q_i=Q_i(q,p,t,\alpha), \space P_i=P_i(q,p,t,\alpha)$$ where $q$ and $p$ represent the set of all $q_i$s and $p_i$s respectively.

From what I leant, The generator corresponding to the infinitesimal Canonical transformations has to be represented in the form: $$F(q,P,t)=\sum q_iP_i +\alpha G(q,P,t) \tag{1}\label{1}$$ Because: Since ICTs have to be infinitesimally close to original coordinates, transformation equations have to be of the form: $$Q_i=q_i+\alpha(something), \space P_i=p_i+\alpha(something) \tag{2}\label{2}$$ (where $\alpha$ is small) and this is made sure by eq(1) because $F=\sum q_iP_i$ corresponds to an identity transformation and hence eq(1) gives identity + something, just like ICT demands in eq(2)

Question: If we consider any general CT given by generator function $F$(need not be a function of $q,P$), depending on a continuous parameter $\alpha$; and we shrink $\alpha$ and neglect the second and higher order terms in $\alpha$, are we guaranteed to get it in the form of eq(1)? (see below for an example)Intuitively I wouldn't think so, because $F$ need not be a function of $q,P$; However, does it mean that when we shrink $\alpha$ the resultant GF doesn't correspond to an ICT? How can it be possible that a finite CT cannot be broken down into many ICTs?

PS: example: if we consider a rotation of coordinate systems by an angle $\alpha$, the GF will be given by: $$F=qPsec(\alpha) -\frac 12 (q^2+P^2)tan(\alpha)$$ for small angles: $$F=qP -\frac 12 (q^2+P^2)\alpha$$ which is clearly in the form of eq(1)

Consider a canonical transformation from $(q,p)$ to $(Q,P)$ depending upon a continuous parameter $\alpha$ such that: $$Q_i=Q_i(q,p,t,\alpha), \space P_i=P_i(q,p,t,\alpha)$$ where $q$ and $p$ represent the set of all $q_i$s and $p_i$s respectively.

From what I leant, The generator corresponding to the infinitesimal Canonical transformations has to be represented in the form: $$F(q,P,t)=\sum q_iP_i +\alpha G(q,P,t). \tag{1}\label{1}$$ Because: Since ICTs have to be infinitesimally close to original coordinates, transformation equations have to be of the form: $$Q_i=q_i+\alpha(something), \space P_i=p_i+\alpha(something) \tag{2}\label{2}$$ (where $\alpha$ is small) and this is made sure by eq(1) because $F=\sum q_iP_i$ corresponds to an identity transformation and hence eq(1) gives identity + something, just like ICT demands in eq(2).

Question: If we consider any general CT given by generator function $F$(need not be a function of $q,P$), depending on a continuous parameter $\alpha$; and we shrink $\alpha$ and neglect the second and higher order terms in $\alpha$, are we guaranteed to get it in the form of eq(1)? (see below for an example)Intuitively I wouldn't think so, because $F$ need not be a function of $q,P$; However, does it mean that when we shrink $\alpha$ the resultant GF doesn't correspond to an ICT? How can it be possible that a finite CT cannot be broken down into many ICTs?

PS: example: if we consider a rotation of coordinate systems by an angle $\alpha$, the GF will be given by: $$F=qP\sec(\alpha) -\frac 12 (q^2+P^2)\tan(\alpha).$$ For small angles: $$F=qP -\frac 12 (q^2+P^2)\alpha$$ which is clearly in the form of eq. (1).