[proofplan] We first record a polynomial growth bound for the Fourier coefficients of the cusp form, obtained by estimating the Fourier coefficients along the horizontal line $\operatorname{Im} z=1/n$. This gives absolute convergence of the Dirichlet series in the half-plane $\operatorname{Re}(s)>k+1$. In that same half-plane, the absolute convergence allows us to interchange the [Fourier series](/page/Fourier%20Series) with the Mellin integral. The remaining one-dimensional integrals are evaluated by the substitution $t=2\pi n y$, giving the factor $(2\pi)^{-s}\Gamma(s)$. [/proofplan] [step:Derive a polynomial bound for the Fourier coefficients] Let $\mathbb{H}=\{z\in\mathbb{C}:\operatorname{Im} z>0\}$, and for $y>0$ define the horizontal segment \begin{align*} \gamma_y:[0,1] &\to \mathbb{H} \\ x &\mapsto x+iy. \end{align*} Since $f$ is a cusp form of weight $k$ for $SL_2(\mathbb{Z})$, it is periodic under $z\mapsto z+1$ and has Fourier expansion \begin{align*} f(x+iy)=\sum_{m=1}^{\infty} a_m e^{2\pi i m x}e^{-2\pi m y} \end{align*} for every $x\in[0,1]$ and $y>0$. Hence Fourier inversion on the unit interval gives, for every $n\in\mathbb{N}$ and every $y>0$, \begin{align*} a_n e^{-2\pi n y} = \int_0^1 f(x+iy)e^{-2\pi i n x}\,d\mathcal{L}^1(x). \end{align*} We claim that there is a constant $C_f>0$ such that \begin{align*} |f(x+iy)|\leq C_f y^{-k} \end{align*} for all $x\in[0,1]$ and all $0<y\leq 1$. This is the standard cusp-form growth estimate at the cusp: applying the modular relation under $S:z\mapsto -1/z$ moves points with small imaginary part to points with large imaginary part, where cuspidality gives exponential decay. Thus, for $n\in\mathbb{N}$, taking $y=1/n$ gives \begin{align*} |a_n| &\leq e^{2\pi n/n}\int_0^1 |f(x+i/n)|\,d\mathcal{L}^1(x) \\ &\leq e^{2\pi} C_f n^k. \end{align*} Define $A_f=e^{2\pi}C_f$. Then \begin{align*} |a_n|\leq A_f n^k \end{align*} for every $n\in\mathbb{N}$. [guided] We need an absolute convergence region in which the later interchange of sum and integral is justified. The Fourier expansion alone gives the exponential factor $e^{-2\pi n y}$, but the coefficients $a_n$ also have to be controlled. For cusp forms, a polynomial bound is enough. For each $y>0$, define the path \begin{align*} \gamma_y:[0,1] &\to \mathbb{H} \\ x &\mapsto x+iy. \end{align*} Because $f$ is invariant under $z\mapsto z+1$, the function $x\mapsto f(x+iy)$ is $1$-periodic. Its Fourier expansion is \begin{align*} f(x+iy)=\sum_{m=1}^{\infty} a_m e^{2\pi i m x}e^{-2\pi m y}. \end{align*} Multiplying by $e^{-2\pi i n x}$ and integrating over $[0,1]$ with respect to one-dimensional Lebesgue measure isolates the $n$th Fourier coefficient: \begin{align*} \int_0^1 f(x+iy)e^{-2\pi i n x}\,d\mathcal{L}^1(x) &= \sum_{m=1}^{\infty} a_m e^{-2\pi m y} \int_0^1 e^{2\pi i(m-n)x}\,d\mathcal{L}^1(x) \\ &=a_n e^{-2\pi n y}. \end{align*} Now use the cusp-form growth estimate near the cusp. Since $f$ has weight $k$ and vanishes at the cusp, there is a constant $C_f>0$ such that \begin{align*} |f(x+iy)|\leq C_f y^{-k} \end{align*} for all $x\in[0,1]$ and all $0<y\leq 1$. Substituting $y=1/n$ into the coefficient formula gives \begin{align*} |a_n|e^{-2\pi} &\leq \int_0^1 |f(x+i/n)|\,d\mathcal{L}^1(x) \\ &\leq \int_0^1 C_f n^k\,d\mathcal{L}^1(x) \\ &=C_f n^k. \end{align*} Therefore, with $A_f=e^{2\pi}C_f$, \begin{align*} |a_n|\leq A_f n^k \end{align*} for every $n\in\mathbb{N}$. [/guided] [/step] [step:Justify the interchange of the Fourier series and the Mellin integral] Let $s\in\mathbb{C}$ and set $\sigma=\operatorname{Re}(s)$. Assume $\sigma>k+1$. For each $n\in\mathbb{N}$, define \begin{align*} g_n:(0,\infty)&\to\mathbb{C} \\ y&\mapsto a_n e^{-2\pi n y}y^{s-1}. \end{align*} Then \begin{align*} \sum_{n=1}^{\infty}\int_0^\infty |g_n(y)|\,d\mathcal{L}^1(y) &= \sum_{n=1}^{\infty}|a_n|\int_0^\infty e^{-2\pi n y}y^{\sigma-1}\,d\mathcal{L}^1(y) \\ &= \Gamma(\sigma)(2\pi)^{-\sigma}\sum_{n=1}^{\infty}|a_n|n^{-\sigma}. \end{align*} Using $|a_n|\leq A_f n^k$, the final series is bounded by \begin{align*} A_f\sum_{n=1}^{\infty}n^{k-\sigma}, \end{align*} which converges because $\sigma>k+1$. Hence the series $\sum_{n=1}^{\infty}g_n$ is absolutely integrable on $(0,\infty)$, and Tonelli's theorem followed by absolute convergence permits termwise integration: \begin{align*} \int_0^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y) = \sum_{n=1}^{\infty} a_n \int_0^\infty e^{-2\pi n y}y^{s-1}\,d\mathcal{L}^1(y). \end{align*} [guided] We now explain why it is legitimate to insert the Fourier expansion into the integral. Let $s\in\mathbb{C}$ and write $\sigma=\operatorname{Re}(s)$. We assume $\sigma>k+1$. For each $n\in\mathbb{N}$, define the measurable function \begin{align*} g_n:(0,\infty)&\to\mathbb{C} \\ y&\mapsto a_n e^{-2\pi n y}y^{s-1}. \end{align*} The absolute value is \begin{align*} |g_n(y)|=|a_n|e^{-2\pi n y}y^{\sigma-1}, \end{align*} because $y>0$ and therefore $|y^{s-1}|=y^{\operatorname{Re}(s)-1}=y^{\sigma-1}$. We estimate the sum of the $L^1$ norms: \begin{align*} \sum_{n=1}^{\infty}\int_0^\infty |g_n(y)|\,d\mathcal{L}^1(y) &= \sum_{n=1}^{\infty}|a_n|\int_0^\infty e^{-2\pi n y}y^{\sigma-1}\,d\mathcal{L}^1(y) \\ &= \Gamma(\sigma)(2\pi)^{-\sigma}\sum_{n=1}^{\infty}|a_n|n^{-\sigma}. \end{align*} The integral evaluation used here is the usual Gamma-integral substitution with real parameter $\sigma>0$. The coefficient bound from the previous step gives \begin{align*} \sum_{n=1}^{\infty}|a_n|n^{-\sigma} \leq A_f\sum_{n=1}^{\infty}n^{k-\sigma}. \end{align*} The $p$-series on the right converges exactly because $k-\sigma<-1$, equivalently $\sigma>k+1$. Thus \begin{align*} \sum_{n=1}^{\infty}\int_0^\infty |g_n(y)|\,d\mathcal{L}^1(y)<\infty. \end{align*} This absolute integrability is the hypothesis needed to interchange the infinite sum and the integral. Therefore, \begin{align*} \int_0^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y) &= \int_0^\infty \sum_{n=1}^{\infty}a_n e^{-2\pi n y}y^{s-1}\,d\mathcal{L}^1(y) \\ &= \sum_{n=1}^{\infty} a_n \int_0^\infty e^{-2\pi n y}y^{s-1}\,d\mathcal{L}^1(y). \end{align*} [/guided] [/step] [step:Evaluate the elementary Mellin integrals] Fix $n\in\mathbb{N}$. Since $\operatorname{Re}(s)>k+1\geq 2$, in particular $\operatorname{Re}(s)>0$, so the Gamma integral converges. Define the change of variables \begin{align*} T_n:(0,\infty)&\to(0,\infty) \\ y&\mapsto 2\pi n y. \end{align*} Under $t=T_n(y)$, one has $d\mathcal{L}^1(y)=(2\pi n)^{-1}d\mathcal{L}^1(t)$ and $y=t/(2\pi n)$. Hence \begin{align*} \int_0^\infty e^{-2\pi n y}y^{s-1}\,d\mathcal{L}^1(y) &= \int_0^\infty e^{-t}\left(\frac{t}{2\pi n}\right)^{s-1}(2\pi n)^{-1}\,d\mathcal{L}^1(t) \\ &= (2\pi n)^{-s}\int_0^\infty e^{-t}t^{s-1}\,d\mathcal{L}^1(t) \\ &= (2\pi n)^{-s}\Gamma(s). \end{align*} [/step] [step:Sum the evaluated integrals to obtain the completed $L$-function] Substituting the integral evaluation into the termwise integral identity gives \begin{align*} \int_0^\infty f(iy)y^{s-1}\,d\mathcal{L}^1(y) &= \sum_{n=1}^{\infty}a_n(2\pi n)^{-s}\Gamma(s) \\ &= (2\pi)^{-s}\Gamma(s)\sum_{n=1}^{\infty}a_n n^{-s} \\ &= (2\pi)^{-s}\Gamma(s)L(f,s) \\ &= \Lambda(f,s). \end{align*} The absolute convergence proved above ensures that every series appearing in this computation converges in the half-plane $\operatorname{Re}(s)>k+1$. This proves the stated Mellin transform formula. [/step]