What means $\int_\Gamma \alpha(\cdot,y)$? I found this map in an exercise of Analysis II of Amann and Escher in page 337 without prior explanation of it meaning
$$f:Y\to\Bbb R,\quad y\mapsto\int_\Gamma\alpha(\cdot,y)\tag1$$
where $\alpha\in\Omega_{(q)}(X\times Y)$ is a $1$-form and $\Gamma$ is a $C^1$ curve in $X$. Here $X\subset\Bbb R^n$ and $Y\subset \Bbb R^m$ are open. Im unable to give a meaning for the expression $\int_\Gamma\alpha(\cdot,y)$.
I know that if $\beta\in\Omega_{(q)}(X)$ and $\gamma:[a,b]\to X$ is a parametrization of $\Gamma$ then
$$\int_\Gamma \beta=\int_a^b\langle \beta(\gamma(t)),\dot\gamma(t)\rangle\, dt$$
What I tried to give a meaning for $\int_\Gamma\alpha(\cdot,y)$: let $\alpha:=\sum_{k=1}^n a_k dx^k+\sum_{j=1}^m b_j dy^j$ where $dx^1,dx^2,\ldots,dx^n,dy^1,\ldots, dy^m$ is the canonical module basis of $\Omega_{(q)}(X\times Y)$, so we can set
$$
\alpha_1:=\sum_{k=1}^n a_k dx^k\in \Omega_{(q)}(X)\quad\text{and}\quad \alpha_2:=\sum_{j=1}^m b_j dy^j\in\Omega_{(q)}(Y)\tag2
$$
Then $\alpha=\alpha_1+\alpha_2$ and we have that
$$
\int_\Gamma\alpha(\cdot,y)=\int_\Gamma\alpha_1+\alpha_2(y)=\int_I\langle\alpha_1(\gamma(t)),\dot\gamma(t)\rangle\, dt+\color{red}{\int_\Gamma\alpha_2(y)}\tag3
$$
But I think that the above manipulation is not totally right. In any case the expression in red seems senseless because $\alpha_2(y)\in T_y^*Y$, where $T_y^*Y$ is the cotangent space around $y$.
Can someone help me to clarify this notation?
 A: I solved the exercise. It is not true that $\alpha_1\in \Omega_{(q)}(X)$ as I supposed because $a_k\notin C^q(X)$, they belong to $C^q(X\times Y)$ as @Ted pointed out in the comment.
The exercise asked to show that $f\in C^q(Y)$ and find an expression for $\nabla f$. 
Let $\alpha:=\sum_{k=1}^n a_k dx^k+\sum_{j=1}^m b_j dy^j$ where $dx^1,\ldots,dx^n,dy^1,\ldots,dy^m$ is the canonical basis of $\Omega_{(q)}(X\times Y)$ for some $a_k,b_j\in C^q(X\times Y)$. Then
$$
\int_\Gamma\alpha(\cdot,y)=\int_{\Gamma\times\{y\}}\alpha\tag1
$$
for the curve $\Gamma\times\{y\}\subset X\times Y$. Then if $\gamma:I\to X$ is a parametrization of $\Gamma$ then
$$
\eta:I\to X\times Y,\quad t\mapsto (\gamma(t),y)\tag2
$$
is a parametrization of $\Gamma\times\{y\}$, hence
$$
f(y)=\int_\Gamma\alpha(\cdot,y)=\int_I\langle\alpha(\eta(t)),\dot\eta(t)\rangle\, dt=\sum_{k=1}^n\int_I a_k(\gamma(t),y)\dot\gamma_k(t)\, dt\tag3
$$
and because $a_k(\gamma(t)),\cdot)\dot\gamma_k(t)\in C^q(Y)$ then its known that $f\in C^q(Y)$. By last its easy to see that
$$
\nabla f(y)=\left(\sum_{k=1}^n\int_I \frac{\partial}{\partial y^1}a_k(\gamma(t),y)\dot\gamma_k(t)\, dt,\ldots,\sum_{k=1}^n\int_I \frac{\partial}{\partial y^m}a_k(\gamma(t),y)\dot\gamma_k(t)\, dt\right)\tag4
$$
or simply
$$
\nabla f=\left(\int_\Gamma\frac{\partial}{\partial y^1}\alpha,\int_\Gamma\frac{\partial}{\partial y^2}\alpha,\ldots,\int_\Gamma\frac{\partial}{\partial y^m}\alpha\right)\tag5
$$
