Find $\displaystyle\int_0^\infty\frac{\sin^4x}{x^4}$ using the fact that $\displaystyle\int_0^\infty\frac{\sin^2x}{x^2} = \frac{\pi}{2}$. The graph of $\dfrac{\sin^4x}{x^4}$ was also given, I tried to integrate by parts repeatedly but it got too messy and confusing. Is there a clever way around this?
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are you trying to find if this is convergent? – Irrational Person Dec 21 '14 at 15:40
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No, trying to find the value of the integral. – user140161 Dec 21 '14 at 15:41
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Wolfram Alpha says $\pi/3$: http://www.wolframalpha.com/input/?i=integral+from+0+to+infty+%5Cfrac%7Bsin%5E4x%7D%7Bx%5E4%7D – Dec 21 '14 at 15:43
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yes I know the answer – user140161 Dec 21 '14 at 15:43
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1I need to do this without complex analysis – user140161 Dec 21 '14 at 15:48
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1One of the answers in Bot's link has the solution you're looking for. – David Mitra Dec 21 '14 at 15:55
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3I have an answer using repeatedly integration by parts, you may have a look here if you're interested. – Venus Dec 21 '14 at 16:10
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I don't understand the last two steps. What happens to the term in $sin^{2}2x$? – user140161 Dec 21 '14 at 17:14
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@user140161 Set $2x\mapsto x$ – Venus Dec 22 '14 at 05:16