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$$\int^{\infty}_{-\infty}\sin\left({\pi}^{4}x^{2}+\frac{1}{x^2}\right) dx$$

This is a problem from the Pi Mu Epsilon Journal, and I'm having great trouble answering it. I've tried some substitutions and any trick I could think of to find some multiple of the integral, but everything has led to a dead end. Perhaps it's not even solvable with methods of real analysis?

Hints and other comments will be much appreciated!

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  • $\begingroup$ Indeed, that's something I used in my attempts but unfortunately I don't see a way of progressing much beyond that. Shouldn't the 2 be on top? $\endgroup$
    – und
    Apr 14, 2013 at 23:38
  • $\begingroup$ I haven't tried that one yet. In fact combining it with some of my other efforts might potentially help. $\endgroup$
    – und
    Apr 14, 2013 at 23:47
  • $\begingroup$ While I was still not sure about whether it converged I put it into Mathematica which gave a non-zero answer. $\endgroup$
    – und
    Apr 14, 2013 at 23:53
  • $\begingroup$ Here is an answer by Maple $ \frac{1}{2}\,{\frac { \left( \cos \left( 2\,{\pi }^{2} \right) +\sin \left( 2 \,{\pi }^{2} \right) \right) \sqrt {2}}{{\pi }^{3/2}}} $ $\endgroup$ Apr 15, 2013 at 0:04
  • $\begingroup$ Here are some techniques. Just use the identity $ \sin(t)=\frac{1}{2 i}(e^{i t}-e^{-i t})$ in order to be able to apply these techniques. $\endgroup$ Apr 15, 2013 at 0:09

2 Answers 2

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By evenness and $\pi x\mapsto u$ we get it is $$\frac{2}{\pi }\int_0^\infty {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)du$$

Now, split at $x=1$, $$\frac{2}{\pi }\int_0^1 {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)du + \frac{2}{\pi }\int_1^\infty {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)du$$ By $u\mapsto u^{-1}$, we get $$\frac{2}{\pi }\int_1^\infty {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)\frac{{du}}{{{u^2}}} + \frac{2}{\pi }\int_1^\infty {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)du$$ or $$\frac{2}{\pi }\int_1^\infty {\sin } \left( {{\pi ^2}\left( {{u^2} + \frac{1}{{{u^2}}}} \right)} \right)\left( {1 + \frac{1}{{{u^2}}}} \right)du$$

Since $u^2+u^{-2}=(u-u^{-1})^2+2$ and $(u-u^{-1})'=1+u^{-2}$ we get $$\frac{2}{\pi }\int_0^\infty {\sin } \left( {{{\left( {\pi x} \right)}^2} + 2{\pi ^2}} \right)dx$$

Now use the sine sum formula and the values of the Fresnel integrals to conclude, the value is: $$\frac{{1 }}{{\sqrt 2{\pi ^{3/2}}}}\left( {\cos 2{\pi ^2} + \sin 2{\pi ^2}} \right)$$

See this for some generalized formulas

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  • $\begingroup$ Expand the sine and you get a cosine and sine Fresnel. Done. $\endgroup$
    – Ron Gordon
    Apr 15, 2013 at 0:14
  • $\begingroup$ @RonGordon Yep. ${}{}{}{}$ $\endgroup$
    – Pedro
    Apr 15, 2013 at 0:15
  • $\begingroup$ Thanks for this! If only I'd known about Fresnel integrals as something to aim for in the manipulations. I assume deriving the results for the Fresnel integrals involves complex analysis? $\endgroup$
    – und
    Apr 15, 2013 at 16:02
  • $\begingroup$ @und I have seen it proven with Laplace's Transform or complex analysis, yes. I never saw a "real" proof, though. Might be worth looking at. By the way, you might accept the answer if you feel like it. $\endgroup$
    – Pedro
    Apr 15, 2013 at 18:43
  • $\begingroup$ @und and Peter: I had asked this same question a few months back: math.stackexchange.com/questions/187729/… $\endgroup$
    – Argon
    Apr 15, 2013 at 19:48
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Not a different answer but rather generalizing the technique Peter Tamaroff used in his answer. Notice the fact that $(u - u^{-1})^2 + 2 = u^2 + u^{-2}$, we can generalize this for any convergent definite integral.

Say if we want to compute: $$ \int^{\infty}_{-\infty} f(x)\,dx, $$ knowing this is convergent. Using the trick we can see the integral is: $$ \int^{\infty}_{-\infty} f(x)\,dx = \int^{\infty}_{-\infty} f\left(x - \frac{a}{x}\right)\,dx \tag{1} $$ Proof: We can make the substitution: $$ u = x - \frac{a}{x}, \quad \text{for some } a>0.$$ Notice $u$ is again from $-\infty$ to $\infty$ for both $x>0$ and $x<0$. Then we have: $$ x = \frac{u + \sqrt{u^2 + 4a}}{2} > 0 \;\text{ or }\; \frac{u - \sqrt{u^2 + 4a}}{2} < 0 $$ hence $$ \frac{dx}{du} = \left\{\begin{aligned} \frac{1}{2}+ \frac{u}{2\sqrt{u^2 + 4a}} \quad \text{when }x>0 \\ \frac{1}{2} - \frac{u}{2\sqrt{u^2 + 4a}} \quad \text{when }x<0 \end{aligned}\right. $$ Hence: $$ \int^{\infty}_{-\infty} f(u)\left(\frac{1}{2}+ \frac{u}{2\sqrt{u^2 + 4a}}\right) du = \int^{\infty}_{0} f\left(x - \frac{a}{x}\right) \,dx \tag{2} $$ and $$ \int^{\infty}_{-\infty} f(u)\left(\frac{1}{2} - \frac{u}{2\sqrt{u^2 + 4a}}\right) du = \int^{0}_{-\infty} f\left(x - \frac{a}{x}\right)\,dx \tag{3} $$ (2)+(3) yields (1).


The integral in OP is $$ \int^{\infty}_{-\infty}\sin\left({\pi}^{4}x^{2}+\frac{1}{x^2}\right) dx = \int^{\infty}_{-\infty}\sin\left({\pi}^4\Big(x - \frac{1}{\pi^2 x}\Big)^2 + 2\pi^2 \right) dx. $$ Now using (1): $$ \int^{\infty}_{-\infty}\sin\left({\pi}^4\Big(x - \frac{1}{\pi^2 x}\Big)^2 + 2\pi^2 \right) dx = \int^{\infty}_{-\infty}\sin ({\pi}^4 x^2 + 2\pi^2) \,dx $$ We can easily see this is the same integral as Peter Tamaroff got in his answer: $$ \frac{2}{\pi }\int_0^\infty {\sin } \left( {{{\left( {\pi x} \right)}^2} + 2{\pi ^2}} \right)dx, $$ and then Fresnel integral kicks in.

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