Well with currents at infinity it's not that easy. Let's first see, where the Biot-Savart law comes from. One should be aware that the fundamental equations are the Maxwell equations in local form, i.e., in terms of partial differential equations.
For magnetostatics the electric and magnetic field separate, and you can treat the magnetic field only. In vacuo the equations read (in SI units)
$$\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}=\mu_0 \vec{j},$$
where ##\vec{j}## is the current density.
To derive the Biot-Savart Law (which almost always is more complicated to evaluate the field than to use the partial differential equations directly) one usually introduces the vector potential. Because of the first equation, there's a vector field ##\vec{A}## such that
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Plugging this in the other equation leads to
$$\vec{\nabla} \times \vec{B} = \vec{\nabla} \times (\vec{\nabla} \times \vec{A}) = \vec{\nabla}(\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=\mu_0 \vec{j}.$$
Now ##\vec{A}## is only determined up to a gradient field (gauge invariance), which implies that we can impose one condition on ##\vec{A}##. Here the most convenient "choice of gauge" is the Coulomb gauge,
$$\vec{\nabla} \cdot \vec{A}=0.$$
Then the equation reads
$$-\Delta \vec{A}=\mu_0 \vec{j}.$$
We know from electrostatics how to solve this for each Cartesian component of ##\vec{A}##:
$$\vec{A}(\vec{x}) = \frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|},$$
which holds if ##\vec{j}## vanishes quickly enough at infinity.
For the most simple example, i.e., an infinitely long wire with constant ##\vec{j}## along it, the integral for ##\vec{A}## does not converge, but for ##\vec{B}## it does
$$\vec{B}(\vec{x})=\vec{\nabla} \times \vec{A}(\vec{x})=\frac{\mu_0}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 x' \vec{j}(\vec{x}') \times \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$
which is Biot-Savart's Law.
Of course, the example for the constant current along an infinite wire is much more simply solved directly using the Maxwell equations. In this case we have
$$\vec{j}(\vec{x})=\vec{e}_z \frac{I}{\pi a^2} \Theta(a-R),$$
where ##(R,\varphi,z)## are the usual cylinder coordinates and ##a## the radius of the cylindrical wire along the ##z##-axis. By symmetry the ansatz for ##\vec{A}## is
$$\vec{A}=A_z(R) \vec{e}_z.$$
Since ##\vec{e}_z=\text{const}## one can use the equation for the Laplacian as applied to a scalar on the component ##A_z##, leading to
$$\Delta A_z(R)=\frac{1}{R} \partial_R (R \partial_R A_z)=-\mu_0 j_z.$$
For ##R<a## we have
$$R \partial_R A_z=C_1 -\mu_0 \frac{I}{2 \pi a^2} R^2$$
and
$$A_z=C_1 \ln(R/a) + C_2 -\mu_0 \frac{I}{4 \pi a^2} R^2.$$
Since ##A_z## should be a smooth function, we must have ##C_1=0##:
$$A_z(R)=C_2-\mu_0 \frac{I}{4 \pi a^2} R^2, \quad 0\leq R<a.$$
The solution for ##R>a## obviously follows for ##I=0##, leading to
$$A_z=C_1 \ln(R/a).$$
To make ##A_z## continuous at ##R=a## we must choose ##C_2=I/(4 \pi)##:
$$A_z(R)=\begin{cases} \frac{\mu_0 I (a^2-R^2)}{4 \pi a^2} &\text{for} \quad 0 \leq R<a,\\
C_1 \ln(R/a) & \text{for} \quad R \geq a. \end{cases}$$
To get ##C_1## we first get
$$\vec{B}=\begin{cases} \frac{I \mu_0 R}{2 \pi a^2} \vec{e}_{\varphi} &\text{for} \quad 0 \leq R <a,\\
-\frac{C_1}{R} \vec{e}_{\varphi} & \text{for} \quad R \geq a. \end{cases}.$$
Since also ##\vec{B}## should be continuous at ##R=a##, we get ##C_1=-I \mu_0/(2 \pi a)## and thus finally
$$\vec{B}=\begin{cases} \frac{\mu_0 I R}{2 \pi a^2} \vec{e}_{\varphi} &\text{for} \quad 0 \leq R <a,\\
\frac{\mu_0 I}{2 \pi R} \vec{e}_{\varphi} & \text{for} \quad R \geq a. \end{cases}.$$
