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Question Number 168338 by infinityaction last updated on 08/Apr/22

Answered by Mathspace last updated on 08/Apr/22

$${\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(ax\right)}{x^2}dxa>0$$$${=}_{ax=t}{\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(t\right)}{{\left(\frac{t}{a}\right)}^2}\frac{dt}{a}$$$$=a{\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}t}{t^2}dtbyparts$$$${\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}t}{t^2}dt={[-\frac{1}{t}{sin}^{2}t]}_0^{\to \mathrm{\infty }}$$$$+{\int }_0^{\mathrm{\infty }}\frac{1}{t}(2sint.cost)dt$$$$=0+{\int }_0^{\mathrm{\infty }}\frac{sin\left(2t\right)}{t}dt(2t=z)$$$$=2{\int }_0^{\mathrm{\infty }}\frac{sinz}{z}\frac{dz}{2}={\int }_0^{\mathrm{\infty }}\frac{sinz}{z}dz=\frac{\pi }{2}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(ax\right)}{x^2}dx=\frac{\pi a}{2}$$$$ifa<0wegetthesameresult$$$$due?tox\to \frac{{sin}^2\left(ax\right)}{x^2}iseven$$$$$$$$$$

Commented by infinityaction last updated on 08/Apr/22

$$thankyousir$$

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