# Why does this fourier transform of the delta train yeild infinity as a result?

For a periodic function denoted as $f(x)=f(x+mx_0)\forall m\in\mathbb{Z}$, I can write the Fourier transform as $$\mathcal{F}(k)=\int_{-\infty}^{+\infty}f(x)\exp{(-ikx)}\mathrm{d}x=\int_{-\infty}^{+\infty}f(x+mx_0)\exp{(-ikx)}\mathrm{d}x$$ If in the above relation the following transformation is assumed $$\lambda=x+mx_0\\ \implies \mathcal{F}(k)=\int_{-\infty}^{+\infty}f(\lambda)\exp{(-ik(\lambda-mx_0))}\mathrm{d}x=\exp{(ikmx_0)}\int_{-\infty}^{+\infty}f(\lambda)\exp{(-ik\lambda)}\mathrm{d}x\\ \implies\exp{(ikmx_0)}=1\implies k=\left(\frac{2\pi}{x_0}\right)n$$ Hence, the transform, for functions periodic in $x_0$ can be written as $$\mathcal{F}(n)=\int_{-\infty}^{+\infty}f(x)\exp{\left(\frac{-i2\pi nx}{x_0}\right)}\mathrm{d}x$$ Now, if I introduce a periodic delta train as shown below $$f(x)=\sum_{p=-\infty}^{+\infty}\delta(x-px_0)\\ \implies \mathcal{F}(n)=\int_{-\infty}^{+\infty}\sum_{p=-\infty}^{+\infty}\delta(x-px_0)\exp{\left(\frac{-i2\pi nx}{x_0}\right)}\mathrm{d}x=\\=\sum_{p=-\infty}^{+\infty}\int_{-\infty}^{+\infty}\delta(x-px_0)\exp{\left(\frac{-i2\pi nx}{x_0}\right)}\mathrm{d}x\\ =\sum_{-\infty}^{+\infty}\exp{(-i2\pi np)}\to\infty$$ But, it is known for a fact that periodic delta trains yeild periodic delta trains in fourier space. Where did I go so radically wrong?

P.S. Any and every help is appreciated, since I am just an engineer and not a mathematician.

After introducing an impluse train$$f(x)=\sum_{p=-\infty}^{+\infty}\delta(x-px_0)$$

You better find the Fourier Series representation of it:

\begin{align} c_k &=\frac{1}{x_0}\int_{-\frac{x_0}{2}}^{\frac{x_0}{2}}\sum_{p=-\infty}^{+\infty}\delta(x-px_0)e^{-i2\pi kt/x_0}\,dx \\ &=\frac{1}{x_0}\sum_{p=-\infty}^{+\infty}\int_{-\frac{x_0}{2}}^{\frac{x_0}{2}}\delta(x-px_0)e^{-i2\pi kt/x_0}\,dx\\&=\frac{1}{x_0}\int_{-\frac{x_0}{2}}^{\frac{x_0}{2}}\delta(x)e^{-i2\pi kt/x_0}\,dx\\&=\frac{1}{x_0}\end{align}

This gives you: $$f(x)=\sum_{p=-\infty}^{+\infty}\delta(x-px_0) \stackrel{F.S.}{=} \frac{1}{x_0}\sum_{k=-\infty}^{+\infty}e^{i2\pi kx/x_o}$$

Then you find the Fourier Transform of $f$:

\begin{align} \mathcal{F}(n)&=\int_{-\infty}^{+\infty}\frac{1}{x_0}\sum_{k=-\infty}^{+\infty}e^{i2\pi kx/x_o}e^{-i2\pi nx}dx \\&=\frac{1}{x_0}\sum_{k=-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{i2\pi kx/x_o}e^{-i2\pi nx}dx \\&=\frac{1}{x_0}\sum_{k=-\infty}^{+\infty}\delta(n-\frac{k}{x_0}) \end{align} which is a scaled train of impulse.

EDIT: To really answer OP's question: I don't view that sum of complex exponential as "quantity". (It doesn't make sense it gives you infinity, right?) To see this result, you can first find the Fourier coefficient, then the Transform.