Integrating functions like $(\sin x)/x$ My books states that the integrals like $\int \frac{\sin x}{x}dx$ and $\int e^{x^2}dx$ exist but they cannot be easily evaluated by elementary functions.I feel it is more because I am unable to evaluate it but can someone please tell me if there is a closed form for them?
 A: Generalize your integral to 
\begin{equation}
I(\alpha)=\int_{0}^{\infty}{dx\, {e}^{-\alpha x}\frac{\sin{x}}{x}}
\end{equation}
Your integral appears as a special case,
\begin{equation}
I(0)=\int_{0}^{\infty}{dx\, \frac{\sin{x}}{x}}
\end{equation} 
Now differentiate $I(\alpha)$ with respect to $\alpha$. Then we get,
\begin{eqnarray}
\frac{\partial{I(\alpha)}}{{\partial\alpha}}=-\int_{0}^{\infty}{dx\, {e}^{-\alpha x}x\times\frac{\sin{x}}{x}}\\
\frac{\partial{I(\alpha)}}{{\partial\alpha}}=-\int_{0}^{\infty}{dx\, {e}^{-\alpha x}\sin{x}} \\
\end{eqnarray} 
Since $\frac{\partial{I(\alpha)}}{{\partial\alpha}}$ is a standard integral and we can use the result $\int{dx\, {e}^{-\alpha x}\sin{x}}= 1/(\alpha^2+1^2)$. It is very easy to derive this result. Hence we have,
\begin{eqnarray}
\frac{\partial{I(\alpha)}}{{\partial\alpha}}=-1/(1+\alpha^2) \\
\end{eqnarray} 
 Integrating the differential equation w.r.t $\alpha$, we get
 \begin{eqnarray}
I(\alpha)=-\tan^{-1}({\alpha})+C \\
\end{eqnarray} 
We know that as $\alpha\to\infty$ the $I(\alpha)$ vanishes. Hence we have $C=\pi/2$, this will give 
\begin{eqnarray}
I(\alpha)=-\tan^{-1}({\alpha})+\frac{\pi}{2}. 
\end{eqnarray} 
Now we can find our integral by simply giving $\alpha=0$ and we get 
\begin{equation}
I(0)=\int_{0}^{\infty}{dx\, \frac{\sin{x}}{x}}=\pi/2
\end{equation} 
This is derivation is meant for those who understand only integration and differentiation. Of Course we can evaluate the integral by contour integral method.
A: Those two functions have no closed form antiderivative with only elementary functions. This is provable. It's sometimes possible to cleanly compute definite integrals involving such functions. For example, 
$$\int_0^\infty \frac{\sin x}{x}\ d x = \frac{\pi}{2}.$$
