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2015-06-06T10:09:54+05:30

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I= \int\limits^{}_{} {\frac{Sin\ x}{x}} \, dx \\\\u=Sin \ x,\ \ \ v'=\frac{1}{x},\ \ \ v=Ln\ x\\\\I=Ln\ x\ Sin\ x- \int\limits^{}_{} {Cos\ x\ Ln\ x} \, dx \\\\=Ln\ x\ Sin\ x-[Cos\ x\ (x\ Ln\ x-x)- \int\limits^{}_{} {(-sin\ x)(x\ Ln\ x-x)} \, dx \\\\=Ln\ x\ Sin\ x-Cos\ x\ (x\ Ln\ x-x)- \int\limits^{}_{} {x\ sin\ x\ Ln\ x} \, dx\\.\ \ \ \ \ + \int\limits^{}_{} {x\ Sin\ x} \, dx \\\\=Lnx\ Sinx-xLnx\ Cosx+x\ Cosx+(-x\ Cosx+Sinx)+I'\\\\I'= - \int\limits^{}_{} {x\ Sinx\ Ln\ x} \, dx \\\\=

This integral does not take a form of a known function or a finite sum of terms in a polynomial.  So we use the Taylor series expansion for Sin x.

Sin\ x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-.....\\\\\frac{Sin\ x}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\frac{x^8}{9!}-.....\\\\ \int\limits^x_0 {\frac{Sin\x}{x}} \, dx =x-\frac{x^3}{3*3!}+\frac{x^5}{5*5!}-\frac{x^7}{7*7!}+\frac{x^9}{9*9!}-....\\\\=\Sigma_{n=0}^{\infty}\ (-1)^{n}\ \frac{x^{2n+1}}{(2n+1)\ *(2n+1)!}

Finally, it is not easy to prove the following using simple calculus methods.  It requires contour integration methods.  The definite integral from 0 to infinity converges :

\int\limits^\infty_0 {\frac{Sin\x}{x}} \, dx =\frac{\pi}{2}

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