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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?

share|cite|improve this question and should get you started. – Kris Jul 23 '12 at 4:27
See this, this, and this, for starters. – J. M. Jul 23 '12 at 4:54

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,

$$\int_0^\infty \frac{\sin x}{x}\ d x = \frac{\pi}{2}.$$

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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.

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