I have to solve the following problem: $$u_t=u_{xx}, x \in \mathbb{R}, t>0$$ $$u(x,0)=f(x)=H(x)=\left\{\begin{matrix} 1, x>0\\ 0, x<0 \end{matrix}\right.$$
I have done the following:
We use the method seperation of variables, $u(x,t)=X(x)T(t)$.
I have found that the eigenfunctions are $X_k(x)=e^{ikx}, \lambda=k^2, k \in \mathbb{R}$
$T_k(t)=e^{-k^2t}$
$$u(x,t)=\frac{1}{2\pi} \int_{-\infty}^{+\infty}{\widetilde{f}(k) e^{ikx} e^{-k^2t}}dk$$
$$u(x,0)=H(x) \Rightarrow H(x)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\widetilde{f}(k)e^{ikx}}dk$$
$$\widetilde{f}(k)=\int_0^{\infty}{e^{-ikx}}dx=\frac{1}{ik}$$
$$u(x,t)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\frac{1}{ik} e^{ikx} e^{-k^2t}}dk$$
Is this correct so far??
How can I calculate the last integral??
EDIT:
I found that the solution of the problem has the following form: $$u(x,t)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\frac{1}{ik} e^{ikx}e^{-k^2t}}dk$$
Then, I derivated this in respect to $x$: $$u_x=\frac{1}{2 \pi} \int_{-\infty}^{+\infty}{\frac{1}{ik} e^{ikx}e^{-k^2t}}dk$$
Then, from the formula: $$\int_{-\infty}^{+\infty}{e^{ikb}e^{-k^2a}}dk=\sqrt{\frac{\pi}{a}}e^{-\frac{b^2}{4a}}$$
we get the following: $$u_x=\frac{1}{2 \pi} \sqrt{\frac{\pi}{t}}e^{-\frac{x^2}{4t}}=\frac{1}{\sqrt{4 \pi t}} e^{-\frac{x^2}{4t}}$$
So to find $u(x,t)$, I have to calculate the integral $\displaystyle{\int u_x dx}$ but with what limits??
Solving an initial value problem for the heat equation by direct calculations of the Fourier transform and its inverse is reasonable only in case it is simpler than applying directly the Poisson formula $$ u(x,t)=\frac{1}{\sqrt{4\pi t}}\int\limits_{-\infty}^{\infty}f(y) e^{-\frac{(x-y)^2}{4t}}dy.\tag{1} $$ Of course, this is not the case with the initial data $f=H$ when $(1)$ turns to $$ \begin{align} u(x,t)=\frac{1}{\sqrt{4\pi t}}\int\limits_{0}^{\infty}e^{-\frac{(x-y)^2}{4t}}dy= \frac{1}{\sqrt{4\pi t}}\int\limits_{0}^{\infty}e^{-\frac{y^2}{4t}}dy+\frac{1}{\sqrt{4\pi t}}\int\limits_{0}^{x}e^{-\frac{y^2}{4t}}dy\\ =\frac{1}{2}+\frac{1}{\sqrt{\pi}}\int\limits_{0}^{x/{\sqrt{4t}}}e^{-s^2}ds= \frac{1}{2}+\frac{1}{2}{\rm erf}\Bigl(\frac{x}{\sqrt{4t}}\Bigr) \end{align} $$ with the Gauss error function ${\rm erf}$.
REMARK added after the OP had been edited.
First, let's take a look at errors in your OP. Generally, an initial value problem for a PDE cannot be solved by separation of variables. In fact, what you are doing might be identified as the method of Fourier transform in the form supposedly inspired by physicists. Though irritating, it looks tolerable for mathematicians' eyes until you write $$ \widetilde{f}(k)=\int_0^{\infty}{e^{-ikx}}dx=\frac{1}{ik}\tag{2} $$ for a divergent integral. Everyone knows that $|e^{ikx}|=1$ when $k\in\mathbb{R}$ which makes $(2)$ a grave mistake. It must be the Fourier transform $\widetilde{f}(k)=F[H]$ instead, being a tempered distribution $\widetilde{f}\in\mathcal{S}'$ in the Schwartz space $\mathcal{S}'$ of linear continuous functionals on the Schwartz space $\mathcal{S}(\mathbb{R})$ of rapidly decreasing functions. The correct version of $(2)$ looks like $$ \widetilde{f}(k)=F[H]=\pi\delta(k)-i\mathcal{P}\frac{1}{k} $$ where functional $\mathcal{P}\frac{1}{k}$, otherwise called a distribution, or a generalized function, operates in accordance with the formula $$ \langle\mathcal{P}\frac{1}{k},\varphi\rangle\overset{\rm def}{=} {\rm Vp}\int\limits_{-\infty}^{+\infty}\frac{\varphi(k)}{k}dk=\lim_{\varepsilon\to +0} \biggl(\int\limits_{-\infty}^{-\varepsilon}+\int\limits_{+\varepsilon}^{+\infty}\biggr) \frac{\varphi(k)}{k}dk\quad\forall\,\varphi\in\mathcal{S}(\mathbb{R}). $$ Sometimes, the more fully detailed notation ${\rm p.v.}\frac{1}{k}$ is employed instead of its shorter, though more widely spread form $\mathcal{P}\frac{1}{k}$ — see, e.g., the section Fourier transform in http://en.wikipedia.org/wiki/Heaviside_step_function
So the correct representation of Fourier transform of the desired solution looks like $$ \widetilde{u}(k,t)=\widetilde{f}(k)e^{-k^2 t}= \Bigl(\pi\delta(k)-i\mathcal{P}\frac{1}{k}\Bigr)e^{-k^2 t}= \pi\delta(k)-ie^{-k^2 t}\mathcal{P}\frac{1}{k} $$ whence follows immediately $$ \begin{align} u(x,t)=\pi F^{-1}[\delta]-F^{-1}\Bigl[e^{-k^2 t}i\mathcal{P}\frac{1}{k}\Bigr]\\ =\frac{1}{2}+\frac{1}{2}{\rm sgn}\ast \Phi=\frac{1}{2}+\frac{1}{2}\int \limits_{-\infty}^{+\infty}{\rm sgn}(y)\Phi(x-y,t)\,dy= \int\limits_{0}^{+\infty}\Phi(x-y,t)\,dy \end{align} $$ with function $$ \Phi(x,t)\overset{\rm def}{=}F^{-1}(e^{-k^2 t})=\frac{1}{2\pi} \int\limits_{-\infty}^{+\infty}e^{ikx}e^{-k^2 t}dk=\frac{e^{-\frac{(x-y)^2}{4t}}}{\sqrt{4\pi t}} $$ Applying the inverse of Fourier transform to tempered distributions requires a certain degree of experience in handling distributions, and is thus less suitable for untrained beginners, to whom the following simplified approach is generally recommended. Namely, observe that $\widetilde{u}(k,t)= \widetilde{f}(k)\widetilde{\Phi}(k,t)$ which by the so-called convolution theorem http://en.wikipedia.org/wiki/Convolution_theorem immediately yields the so-called Poisson formula $u=f\ast\Phi$, i.e., formula $(1)$. This Poisson formula $(1)$ can be found in almost any textbook on PDE, e.g., in that by Evans (see p. 47 of 2nd ed.).
Second, a solution (no matter classical or weak) represented by formula $(1)$ is unique in the class of functions increasing at infinity not faster than $e^{Mx^2}$, i.e., satisfying the inequality $$ |u(x,t)|\leqslant Ce^{Mx^2}\quad\forall\,(x,t)\in\mathbb{R}\times [0,T] $$ with some positive constants $C,M$ depending only on the prescribed value of $T>0$. This is called the uniqueness theorem of A. Tikhonov http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=6410&option_lang=eng That is why the bounded function $$ u(x,t)=\frac{1}{2}+\frac{1}{2}{\rm erf}\Bigl(\frac{x}{\sqrt{4t}}\Bigr), \quad 0<u(x,t)<1,\tag{3} $$ will be the only solution to the heat equation with the intial data $u(x,0)=f(x)=H(x)$. A most thoroughly studied special function ${\rm erf}$ can be represented otherwise, but cannot be simplified. So the only correct answer to your question is given by $(3)$, and no one can ever reduce it to a more simple form.