I don't have enough reputation to comment so I am leaving this as an answer. You are integrating over the square of side-length $2\pi$, a reasonable first step (that might make the minimum more easy to handle) is to shift to a coordinate system that is rotated at 45 degrees to the one you're currently using, something like

\begin{align} u &= \frac{1}{\sqrt{2}}(x+y)\\ v &= \frac{1}{\sqrt{2}}(x-y), \end{align}

the limits of integration are a bit annoying with this subsitution, one of them is simple you just go from the origin to the top right corner of the square so the interval is $[0, 2\pi\sqrt{2}]$. The second limit has to between the edges of the square, orthogonal to the diagonal. Its simplest to break the first integral when we reach the center of the square so we can express this relatively simply

\begin{align} I = &\int_{0}^{\pi\sqrt{2}}\int_{-v}^v \min\{v\sqrt{2}, 2\pi-v\sqrt{2}\} e^{a\cos\left(\frac{u+v}{\sqrt{2}}\right)} e^{a\cos\left(\frac{u-v}{\sqrt{2}}\right)} du dv \\&+ \int_{\pi\sqrt{2}}^{2\pi\sqrt{2}}\int_{v-2\pi\sqrt{2}}^{2\pi\sqrt{2}-v} \min\{v\sqrt{2}, 2\pi-v\sqrt{2}\} e^{a\cos\left(\frac{u+v}{\sqrt{2}}\right)} e^{a\cos\left(\frac{u-v}{\sqrt{2}}\right)} du dv. \end{align}

At this point you can split up the $v$ integrals further to get rid of the minimums and (hopefully!) make an integral you can deal with.

I don't have enough reputation to comment so I am leaving this as an answer. You are integrating over the square of side-length $2\pi$, a reasonable first step (that might make the minimum more easy to handle) is to shift to a coordinate system that is rotated at 45 degrees to the one you're currently using, something like

\begin{align} u &= \frac{1}{\sqrt{2}}(x+y)\\ v &= \frac{1}{\sqrt{2}}(x-y), \end{align}

the limits of integration are a bit annoying with this subsitution, one of them is simple you just go from the origin to the top right corner of the square so the interval is $[0, 2\pi\sqrt{2}]$. The second limit has to between the edges of the square, orthogonal to the diagonal. Its simplest to break the first integral when we reach the center of the square so we can express this relatively simply

\begin{align} I = &\int_{0}^{\pi\sqrt{2}}\int_{-v}^v \min\{v\sqrt{2}, 2\pi-v\sqrt{2}\} e^{a\cos\left(\frac{u+v}{\sqrt{2}}\right)} e^{a\cos\left(\frac{u-v}{\sqrt{2}}\right)} \, du \, dv \\&+ \int_{\pi\sqrt{2}}^{2\pi\sqrt{2}}\int_{v-2\pi\sqrt{2}}^{2\pi\sqrt{2}-v} \min\{v\sqrt{2}, 2\pi-v\sqrt{2}\} e^{a\cos\left(\frac{u+v}{\sqrt{2}}\right)} e^{a\cos\left(\frac{u-v}{\sqrt{2}}\right)} \, du \, dv. \end{align}

At this point you can split up the $v$ integrals further to get rid of the minimums and (hopefully!) make an integral you can deal with.

I don't have enough reputation to comment so I am leaving this as an answer. You are integrating over the square of side-length $2\pi$, a reasonable first step (that might make the minimum more easy to handle) is to shift to a coordinate system that is rotated at 45 degrees to the one you're currently using, something like

\begin{align} u &= \frac{1}{\sqrt{2}}(x+y)\\ v &= \frac{1}{\sqrt{2}}(x-y), \end{align}

the limits of integration are a bit annoying with this subsitution, one of them is simple you just go from the origin to the top right corner of the square so the interval is $[0, 2\pi\sqrt{2}]$. The second limit has to go from the diagonal to the edges of the square. Its simplest to break the first integral when we reach the center of the square so we can express this relatively simply

\begin{align} I = &\int_{0}^{\pi\sqrt{2}}\int_{-v}^v \min\{v\sqrt{2}, 2\pi-v\sqrt{2}\} e^{a\cos\left(\frac{u+v}{\sqrt{2}}\right)} e^{a\cos\left(\frac{u-v}{\sqrt{2}}\right)} du dv \\&+ \int_{\pi\sqrt{2}}^{2\pi\sqrt{2}}\int_{v-2\pi\sqrt{2}}^{2\pi\sqrt{2}-v} \min\{v\sqrt{2}, 2\pi-v\sqrt{2}\} e^{a\cos\left(\frac{u+v}{\sqrt{2}}\right)} e^{a\cos\left(\frac{u-v}{\sqrt{2}}\right)} du dv. \end{align}

At this point you can split up the $v$ integrals further to get rid of the minimums and (hopefully!) make an integral you can deal with.

I don't have enough reputation to comment so I am leaving this as an answer. You are integrating over the square of side-length $2\pi$, a reasonable first step (that might make the minimum more easy to handle) is to shift to a coordinate system that is rotated at 45 degrees to the one you're currently using, something like

\begin{align} u &= \frac{1}{\sqrt{2}}(x+y)\\ v &= \frac{1}{\sqrt{2}}(x-y), \end{align}

the limits of integration are a bit annoying with this subsitution, one of them is simple you just go from the origin to the top right corner of the square so the interval is $[0, 2\pi\sqrt{2}]$. The second limit has to between the edges of the square, orthogonal to the diagonal. Its simplest to break the first integral when we reach the center of the square so we can express this relatively simply

\begin{align} I = &\int_{0}^{\pi\sqrt{2}}\int_{-v}^v \min\{v\sqrt{2}, 2\pi-v\sqrt{2}\} e^{a\cos\left(\frac{u+v}{\sqrt{2}}\right)} e^{a\cos\left(\frac{u-v}{\sqrt{2}}\right)} du dv \\&+ \int_{\pi\sqrt{2}}^{2\pi\sqrt{2}}\int_{v-2\pi\sqrt{2}}^{2\pi\sqrt{2}-v} \min\{v\sqrt{2}, 2\pi-v\sqrt{2}\} e^{a\cos\left(\frac{u+v}{\sqrt{2}}\right)} e^{a\cos\left(\frac{u-v}{\sqrt{2}}\right)} du dv. \end{align}

At this point you can split up the $v$ integrals further to get rid of the minimums and (hopefully!) make an integral you can deal with.