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?
Integrating functions like $(\sin x)/x$
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0@tired : I'm not sure that the given link is a duplicate. I don't think the OP asked for definite integrals, but rather for closed-form primitive functions. In my opinion the links given by J. M. are more relevant. There also is [this](http://math.stackexchange.com/q/830833) answer. – 2016-03-03
2 Answers
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.
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,
$\int_0^\infty \frac{\sin x}{x}\ d x = \frac{\pi}{2}.$