Mathematics

# Evaluate $\displaystyle\int_0^\infty{\sin{x}dx}$

1

##### SOLUTION
Let us first evaluate;
$\displaystyle I=\int{e^{\displaystyle-sx}\sin{x}dx}$ and $\displaystyle J=\int{e^{\displaystyle-sx}\cos{x}dx}$
Using integer by parts, we get
$I=-e^{\displaystyle-sx}\cos{x}-sJ$ (i)
$J=e^{\displaystyle-sx}\sin{x}+sI$ (ii)
Subtracting equations (i) and (ii), we get
$\displaystyle I=-e^{\displaystyle-sx}\left|\frac{\cos{x}+s\sin{x}}{1+s^2}\right|$
$\displaystyle\implies J=e^{\displaystyle-sx}\left|\sin{x}-\frac{s^2}{s^2+1}\sin{x}-\frac{s}{s^2+1}\cos{x}\right|$
$\displaystyle e^{\displaystyle-sx}\left|\frac{\sin{x}-s\cos{x}}{1+s^2}\right|$
Thus, $\displaystyle\int_0^\infty{e^{\displaystyle-sx}\sin{x}dx}=\frac{1}{s^2+1}$
$\displaystyle\int_0^\infty{e^{\displaystyle-sx}\cos{x}dx}=\frac{s}{s^2+1}$
Now, $\displaystyle\int_0^\infty{\sin{x}dx}=\lim_{s\rightarrow0}{\int_0^\infty{e^{\displaystyle-sx}\sin{x}dx}}=\lim_{s\rightarrow0}{\frac{1}{s^2+1}}=1$

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One Word Medium Published on 17th 09, 2020
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