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I really wonder how I can prove the following integrals.

$$\int_0^\infty \sin ax^2\cos 2bx\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}\left(\cos \frac{b^2}{a}-\sin\frac{b^2}{a}\right)$$

and

$$\int_0^\infty \cos ax^2\cos 2bx\, dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}\left(\cos \frac{b^2}{a}+\sin\frac{b^2}{a}\right)$$


I tried $\sin ax^2=\Im(e^{iax^2})$ and $\cos ax^2=\Re(e^{iax^2})$ then I used by parts method but I failed. Obviously tangent half-angle substitution doesn't work. I'm quite sure if we can calculate one of them, the similar technique can be used to calculate the other. Could anyone here please help me to calculate the integrals preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

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Compute them together:

Add $i$ times the first integral and the second, to obtain

$$\int_0^\infty e^{iax^2}\cos (2bx)\,dx.$$

The integrand is even, so that is

$$\frac{1}{2}\int_{-\infty}^\infty e^{iax^2}\cos (2bx)\,dx.$$

$\sin$ is odd, so the integral of $e^{iax^2}\sin (2bx)$ vanishes, hence we get

$$\frac{1}{2} \int_{-\infty}^\infty e^{iax^2}e^{2bix}\,dx.$$

Complete the square in the exponent to obtain

$$\frac{1}{2} e^{-ib^2/a} \int_{-\infty}^\infty e^{i(\sqrt{a}x + b/\sqrt{a})^2}\,dx.$$

Substitute $y = \sqrt{a}x + b/\sqrt{a}$, and you obtain

$$\frac{1}{2\sqrt{a}} e^{-ib^2/a} \int_{-\infty}^\infty e^{iy^2}\,dy.$$

The last integral has standard Fresnel integrals as real and imaginary parts, so its value is $(1+i)\sqrt{\frac{\pi}{2}}$, and altogether we get

$$\int_0^\infty \cos (ax^2)\cos (2bx)\,dx + i\int_0^\infty \sin (ax^2)\cos (2bx)\,dx = \frac{1}{2\sqrt{a}}e^{-ib^2/a}\sqrt{\frac{\pi}{2}}(1+i).$$

Separate the right hand side into real and imaginary part.

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  • $\begingroup$ I can calculate it from the third line. It never crosses to mind using that way. +1 $\endgroup$ – Anastasiya-Romanova 秀 Jul 25 '14 at 16:20
  • $\begingroup$ I forget to say thank you. Thanks! Brilliant answer!! (>‿◠)✌ $\endgroup$ – Anastasiya-Romanova 秀 Jul 25 '14 at 16:27
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Notice that $\mathbb{sin}A \mathbb{cos}B = \frac{1}{2}(\mathbb{sin}(A+B)+\mathbb{sin}(A-B))$. Also notice that $ax^2+bx=(\sqrt{a}x+\frac{b}{2\sqrt{a}})^2-\frac{b^2}{4a}$ and $ax^2-bx=(\sqrt{a}x-\frac{b}{2\sqrt{a}})^2-\frac{b^2}{4a}$. You will get two Fresnel integrals in which you will have to make the changes of variables $\sqrt{a}x+\frac{b}{2\sqrt{a}}=y$ and $\sqrt{a}x-\frac{b}{2\sqrt{a}}=y$, respectively.

For the second integral use that $\mathbb{cos}A \mathbb{cos}B = \frac{1}{2}(\mathbb{cos}(A+B)+\mathbb{cos}(A-B))$ and redo the computations above. (Of course, I am assuming $a>0$.)

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  • $\begingroup$ Thank you for your answer. This brilliant answer too. +1 (>‿◠)✌ $\endgroup$ – Anastasiya-Romanova 秀 Jul 25 '14 at 16:28
  • $\begingroup$ The one who downvoted my answer, could you also explain your behaviour, please? $\endgroup$ – Alex M. Aug 9 '14 at 17:07

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