[tex]M_X(t)=\mathbb E(e^{Xt})[/tex]
[tex]M_X(t)=\mathbb E\left(1+Xt+\dfrac{t^2}{2!}X^2+\dfrac{t^3}{3!}X^3+\cdots\right)[/tex]
[tex]M_X(t)=\mathbb E(1)+t\mathbb E(X)+\dfrac{t^2}{2!}\mathbb E(X^2)+\dfrac{t^3}{3!}\mathbb E(X^3)+\cdots[/tex]
[tex]M_X(t)=1+t+t^2+t^3+\cdots[/tex]
[tex]M_X(t)=\displaystyle\sum_{k\ge0}t^k=\frac1{1-t}[/tex]
provided that [tex]|t|<1[/tex].
Similarly,
[tex]\varphi_X(t)=\mathbb E(e^{iXt})[/tex]
[tex]\varphi_X(t)=1+it+(it)^2+(it)^3+\cdots[/tex]
[tex]\varphi_X(t)=(1-t^2+t^4-t^6+\cdots)+it(1-t^2+t^4-t^6+\cdots)[/tex]
[tex]\varphi_X(t)=(1+it)(1-t^2+t^4-t^6+\cdots)[/tex]
[tex]\varphi_X(t)=\dfrac{1+it}{1+t^2}=\dfrac1{1-it}[/tex]
You can find the CDF/PDF using any of the various inversion formulas. One way would be to compute
[tex]F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{e^{itx}\varphi_X(-t)-e^{-itx}\varphi_X(t)}{it}\,\mathrm dt[/tex]
The integral can be rewritten as
[tex]\displaystyle\int_0^\infty\frac{2i\sin(tx)-2it\cos(tx)}{it(1+t^2)}\,\mathrm dt[/tex]
so that
[tex]F_X(x)=\displaystyle\frac12+\frac1{2\pi}\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt[/tex]
There are lots of ways to compute this integral. For instance, you can take the Laplace transform with respect to [tex]x[/tex], which gives
[tex]\displaystyle\mathcal L_s\left\{\int_0^\infty\frac{\sin(tx)-t\cos(tx)}{t(1+t^2)}\,\mathrm dt\right\}=\int_0^\infty\frac{1-s}{(1+t^2)(s^2+t^2)}\,\mathrm dt[/tex]
[tex]=\displaystyle\frac{\pi(1-s)}{2s(1+s)}[/tex]
and taking the inverse transform returns
[tex]F_X(x)=\dfrac12+\dfrac1\pi\left(\dfrac\pi2-\pi e^{-x}\right)=1-e^{-x}[/tex]
which describes an exponential distribution with parameter [tex]\lambda=1[/tex].
