Questions regarding $\int_0^\infty\frac{\sin^2x}{x^2}dx$

Asked Dec 24 '15 at 17:08

Active Dec 24 '15 at 17:41

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question: $\int_0^\infty\frac{\sin^2x}{x^2}dx$ is equal to

(A)$\int_0^\infty\frac{\sin x}{x}dx$

(B)$\int_0^\infty\frac{\cos x}{x}dx$

(C)$\int_0^\infty\frac{\cos^2x}{x^2}dx$

(D)None of the above

I tried splitting the integral as $$\int_0^\infty\frac{1-\cos^2x}{x^2}dx=\int_0^\infty\frac{1}{x^2}dx-\int_0^\infty\frac{\cos^2x}{x^2}dx$$ But for the first integrand, the function is not defined when $x=0$. Then I tried using $$\sin^2x=\frac{1-\cos 2x}{2}$$ but that did not work as well.

edited Jun 12 '20 at 10:38

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asked Dec 24 '15 at 17:08

the right answer is only $A$ – Dr. Sonnhard Graubner Dec 24 '15 at 17:17
That's correct. How did you solve it? – Dec 24 '15 at 17:19
see here http://hubpages.com/education/How-to-Integrate-sinxx-and-cosxx – Dr. Sonnhard Graubner Dec 24 '15 at 17:22
and here http://math.stackexchange.com/questions/141695/how-to-calculate-the-integral-of-sin2x-x2 – Dr. Sonnhard Graubner Dec 24 '15 at 17:24
see this link http://math.stackexchange.com/questions/13344/proof-of-int-0-infty-left-frac-sin-xx-right2-mathrm-dx-frac-pi2 – Mosk Dec 24 '15 at 17:28
https://math.stackexchange.com/q/2508827 – Guy Fsone Jan 02 '18 at 09:31

Questions regarding $\int_0^\infty\frac{\sin^2x}{x^2}dx$

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