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

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!

  • $\begingroup$ 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 '13 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 '13 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 '13 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}}} $ $\endgroup$ – Mhenni Benghorbal Apr 15 '13 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. $\endgroup$ – Mhenni Benghorbal Apr 15 '13 at 0:09

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

  • $\begingroup$ Expand the sine and you get a cosine and sine Fresnel. Done. $\endgroup$ – Ron Gordon Apr 15 '13 at 0:14
  • $\begingroup$ @RonGordon Yep. ${}{}{}{}$ $\endgroup$ – Pedro Tamaroff Apr 15 '13 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 '13 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 Tamaroff Apr 15 '13 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 '13 at 19:48

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.