Mathematics

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


ANSWER

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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