[proofplan]
We first identify the completed $L$-function with the Mellin transform of $f$ along the positive imaginary axis in a right half-plane. We then split the Mellin integral at $1$, transform the interval $(0,1)$ by $y = 1/t$, and use the modular relation under $z \mapsto -1/z$. The transformed expression is an integral over $(1,\infty)$ whose exponential decay makes it entire in $s$, and the same expression gives the functional equation by replacing $s$ with $k-s$.
[/proofplan]
[step:Reduce the odd weight case to the zero form]
If $k$ is odd, then $f = 0$. Indeed, since $-I \in \operatorname{SL}_2(\mathbb{Z})$, the weight $k$ transformation law gives
\begin{align*}
f(z) = (-1)^k f(z)
\end{align*}
for every $z \in \mathbb{H}$. Hence $f(z) = -f(z)$ for all $z \in \mathbb{H}$, so $f = 0$. In that case $L(f,s)=0$ and $\Lambda(f,s)=0$, so the asserted continuation and functional equation hold. We therefore assume for the rest of the proof that $k$ is even.
[/step]
[step:Identify $\Lambda(f,s)$ with a Mellin integral]
Since $f$ is a cusp form, its Fourier expansion on the imaginary axis is
\begin{align*}
f(iy) = \sum_{n=1}^{\infty} a_n e^{-2\pi n y},
\qquad y \in (0,\infty),
\end{align*}
and the Fourier coefficients have at most polynomial growth. Thus for $\operatorname{Re}(s)$ sufficiently large, the following integral is absolutely convergent and termwise integration is justified. Define
\begin{align*}
M:\{s \in \mathbb{C} : \operatorname{Re}(s) \text{ sufficiently large}\} &\to \mathbb{C} \\
s &\mapsto \int_0^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y).
\end{align*}
Then
\begin{align*}
M(s)
&= \sum_{n=1}^{\infty} a_n \int_0^\infty e^{-2\pi n y}y^{s-1}\,d\mathcal{L}^1(y).
\end{align*}
For each $n \in \mathbb{N}$, using the substitution $u = 2\pi n y$, so that $d\mathcal{L}^1(u)=2\pi n\,d\mathcal{L}^1(y)$, gives
\begin{align*}
\int_0^\infty e^{-2\pi n y}y^{s-1}\,d\mathcal{L}^1(y)
&= (2\pi n)^{-s}\int_0^\infty e^{-u}u^{s-1}\,d\mathcal{L}^1(u) \\
&= (2\pi n)^{-s}\Gamma(s).
\end{align*}
Therefore
\begin{align*}
M(s)
&= (2\pi)^{-s}\Gamma(s)\sum_{n=1}^{\infty} a_n n^{-s}
= \Lambda(f,s)
\end{align*}
in that right half-plane.
[/step]
[step:Transform the small part of the Mellin integral using modularity]
For $\operatorname{Re}(s)$ sufficiently large, split
\begin{align*}
M(s)
= \int_1^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y)
+ \int_0^1 f(iy)y^{s-1}\,d\mathcal{L}^1(y).
\end{align*}
In the second integral, use the substitution $y = 1/t$, which maps $t \in (1,\infty)$ onto $y \in (0,1)$ and satisfies
\begin{align*}
d\mathcal{L}^1(y)=t^{-2}\,d\mathcal{L}^1(t)
\end{align*}
after reversing the orientation of the interval. Hence
\begin{align*}
\int_0^1 f(iy)y^{s-1}\,d\mathcal{L}^1(y)
&= \int_1^\infty f(i/t)t^{-s-1}\,d\mathcal{L}^1(t).
\end{align*}
The matrix
\begin{align*}
S=
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\in \operatorname{SL}_2(\mathbb{Z})
\end{align*}
acts by $S z = -1/z$. The weight $k$ transformation law gives
\begin{align*}
f(-1/z)=z^k f(z),
\qquad z \in \mathbb{H}.
\end{align*}
Taking $z=it$ with $t \in (1,\infty)$ yields
\begin{align*}
f(i/t)=f(-1/(it))=(it)^k f(it)=i^k t^k f(it).
\end{align*}
Therefore
\begin{align*}
\int_0^1 f(iy)y^{s-1}\,d\mathcal{L}^1(y)
&= i^k \int_1^\infty f(it)t^{k-s-1}\,d\mathcal{L}^1(t).
\end{align*}
Thus, in the initial half-plane,
\begin{align*}
\Lambda(f,s)
= \int_1^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y)
+ i^k \int_1^\infty f(iy)y^{k-s-1}\,d\mathcal{L}^1(y).
\end{align*}
[guided]
The point of splitting at $1$ is that the modular transformation $z \mapsto -1/z$ exchanges the small imaginary values $iy$ with large imaginary values. We first write
\begin{align*}
M(s)
= \int_1^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y)
+ \int_0^1 f(iy)y^{s-1}\,d\mathcal{L}^1(y).
\end{align*}
The first integral is already over the region where cusp forms decay exponentially. For the second integral, set $y = 1/t$. As $y$ runs from $0$ to $1$, the variable $t$ runs from $\infty$ to $1$, and
\begin{align*}
d\mathcal{L}^1(y)=t^{-2}\,d\mathcal{L}^1(t)
\end{align*}
after reversing the limits. Hence
\begin{align*}
\int_0^1 f(iy)y^{s-1}\,d\mathcal{L}^1(y)
&= \int_1^\infty f(i/t)t^{-s-1}\,d\mathcal{L}^1(t).
\end{align*}
Now we use modularity. The element
\begin{align*}
S=
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\end{align*}
belongs to $\operatorname{SL}_2(\mathbb{Z})$ and sends $z$ to $-1/z$. Since $f$ has weight $k$,
\begin{align*}
f(-1/z)=z^k f(z).
\end{align*}
Putting $z=it$ gives $-1/(it)=i/t$, so
\begin{align*}
f(i/t)=f(-1/(it))=(it)^k f(it)=i^k t^k f(it).
\end{align*}
Substituting this into the transformed integral gives
\begin{align*}
\int_0^1 f(iy)y^{s-1}\,d\mathcal{L}^1(y)
&= i^k \int_1^\infty f(it)t^{k-s-1}\,d\mathcal{L}^1(t).
\end{align*}
Renaming the integration variable $t$ as $y$ gives the rewritten Mellin formula
\begin{align*}
\Lambda(f,s)
= \int_1^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y)
+ i^k \int_1^\infty f(iy)y^{k-s-1}\,d\mathcal{L}^1(y).
\end{align*}
[/guided]
[/step]
[step:Use exponential decay to continue the rewritten expression entire]
Because $f$ is cuspidal at infinity, there exist constants $C>0$ and $\alpha>0$ such that
\begin{align*}
|f(iy)| \leq C e^{-\alpha y},
\qquad y \geq 1.
\end{align*}
Define
\begin{align*}
F:\mathbb{C} &\to \mathbb{C} \\
s &\mapsto
\int_1^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y)
+ i^k \int_1^\infty f(iy)y^{k-s-1}\,d\mathcal{L}^1(y).
\end{align*}
Let $K \subset \mathbb{C}$ be compact, and choose [real numbers](/page/Real%20Numbers) $A,B$ such that
\begin{align*}
A \leq \operatorname{Re}(s) \leq B
\end{align*}
for every $s \in K$. For $y \geq 1$ and $s \in K$,
\begin{align*}
|f(iy)y^{s-1}| &\leq C e^{-\alpha y} y^{B-1}, \\
|f(iy)y^{k-s-1}| &\leq C e^{-\alpha y} y^{k-A-1}.
\end{align*}
Both dominating functions are integrable on $(1,\infty)$ with respect to $\mathcal{L}^1$. The same estimates with extra powers of $\log y$ dominate all complex derivatives with respect to $s$. Hence both integrals defining $F(s)$ converge locally uniformly in $s$ and may be differentiated term by term. Therefore $F$ is entire.
The previous step proved $F(s)=\Lambda(f,s)$ on a nonempty right half-plane. Thus $F$ is the [analytic continuation](/page/Analytic%20Continuation) of $\Lambda(f,s)$ to an entire function on $\mathbb{C}$.
[/step]
[step:Read the functional equation from the continued integral formula]
For every $s \in \mathbb{C}$, the continued function is
\begin{align*}
F(s)
= \int_1^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y)
+ i^k \int_1^\infty f(iy)y^{k-s-1}\,d\mathcal{L}^1(y).
\end{align*}
Replacing $s$ by $k-s$ gives
\begin{align*}
F(k-s)
= \int_1^\infty f(iy)y^{k-s-1}\,d\mathcal{L}^1(y)
+ i^k \int_1^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y).
\end{align*}
Since $k$ is even, $i^{2k}=1$. Multiplying the last identity by $i^k$ yields
\begin{align*}
i^k F(k-s)
&= i^k \int_1^\infty f(iy)y^{k-s-1}\,d\mathcal{L}^1(y)
+ i^{2k} \int_1^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y) \\
&= i^k \int_1^\infty f(iy)y^{k-s-1}\,d\mathcal{L}^1(y)
+ \int_1^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y) \\
&= F(s).
\end{align*}
Since $F$ is the entire continuation of $\Lambda(f,s)$, this proves
\begin{align*}
\Lambda(f,s)=i^k\Lambda(f,k-s)
\end{align*}
for all $s \in \mathbb{C}$.
[/step]
