10
$\begingroup$

I tried this: $$\int_0^{\infty} \cos\left(\frac{a^2}{x^2}-b^2x^2\right)\,dx=\Re\left(\int_0^{\infty} e^{-ib^2x^2+ia^2/x^2}\,dx\right)=\Re\left(\int_0^{\infty} e^{-\left(ib^2x^2+i^3a^2/x^2\right)}\,dx\right)$$ Sometime back, I stumbled upon the following result: $$\int_0^{\infty} e^{-\left(p^2x^2+m^2/x^2\right)}\,dx=\frac{\sqrt{\pi}}{2p}e^{-2pm}$$ Replacing $p$ with $i^{1/2}b$ and $m$ with $i^{3/2}a$, I get: $$\int_0^{\infty} \cos\left(\frac{a^2}{x^2}-b^2x^2\right)\,dx=\Re\left(\frac{1}{2b}\sqrt{\frac{\pi}{i}}e^{2ab}\right)$$ But this is supposed to be incorrect and I do not see where did I go wrong.

Any help is greatly appreciated. Thanks!

(I do know that this is easily doable using contour integration but I would like to know what's wrong with above)

$\endgroup$
  • $\begingroup$ Supposed to be incorrect? Did you actually find the real part of that expression and check? $\endgroup$ – M. Vinay Jul 13 '14 at 3:48
  • $\begingroup$ @M.Vinay: Yes, I did. $\endgroup$ – Pranav Arora Jul 13 '14 at 3:48
  • $\begingroup$ Make sure you simlified the last result correctly. $\endgroup$ – Mhenni Benghorbal Jul 13 '14 at 4:02
  • $\begingroup$ You should have a $-2ab$ in the exponent I believe $\endgroup$ – ClassicStyle Jul 13 '14 at 4:19
  • $\begingroup$ Just checked. With that fix you will get the correct answer $\endgroup$ – ClassicStyle Jul 13 '14 at 4:21
5
$\begingroup$

The confusion arises from the fact that you're assuming the formula $$\int_0^{\infty} e^{-\left(p^2x^2+m^2/x^2\right)}\,dx=\frac{\sqrt{\pi}}{2p}e^{-2pm} $$ is valid if $p^{2}$ is imaginary and positive and $m^{2}$ is imaginary and negative. But that formula is usually derived under the condition that $p$ and $m$ are real positive parameters.

But by integrating on the complex plane, you can see how the two integrals are related.

Let $ \displaystyle f(z) = e^{i b^{2}z^{2}} e^{-ia^{2}/z^{2}}$ and integrate around a wedge/sector of radius $R$ that makes an angle of $ \frac{\pi}{4}$ with the positive real axis and is indented around the essential singularity at the origin.

Along the arc of the wedge, $ \displaystyle |e^{-ia^{2}/z^{2}}|= |e^{-i a^{2}/(R^{2}e^{2it})}| =e^{-a^{2} \sin 2t/R^{2}} \le 1 $ since $ \displaystyle 0 \le t \le \frac{\pi}{4}$.

Therefore,

$$\begin{align} \Big| \int_{0}^{\pi /4} f(Re^{it}) \ i Re^{it} \ dt \Big| &\le R \int_{0}^{\pi /4} e^{-b^{2} R^{2} \sin 2t} \ dt \\ &\le R \int_{0}^{\pi /4} e^{-b^{2} R^{2} \frac{4}{\pi} t } \ dt \ \ \text{(Jordan's inequality)} \\ &= \frac{\pi}{4} \frac{1}{b^{2}R} \Big( 1-e^{-b^{2}R^{2}}\Big) \to 0 \ \text{as} \ R \to \infty .\end{align}$$

A very similar argument shows that the integral also vanishes along the quarter-circle indentation around the origin as the radius of the quarter-circle goes to $0$.

Therefore, since $f(z)$ is analytic inside and on the contour,

$$ \int_{0}^{\infty} f(x) \ dx - \int_{0}^{\infty} f(te^{i \pi /4}) e^{i \pi /4} \ dt =0$$

which implies

$$ \begin{align}\int_{0}^{\infty}e^{i b^{2}x^{2}} e^{-ia^{2}/x^{2}} \ dx &= e^{i \pi /4} \int_{0}^{\infty} e^{-b^{2}t^{2}} e^{-a^{2}/t^{2}} \ dt \\ &= e^{i \pi /4} \frac{\sqrt{\pi}}{2b} e^{-2 ab}. \end{align}$$

And equating the real parts on both sides of the equation,

$$ \begin{align} \int_{0}^{\infty} \cos \left(b^{2}x^{2} -\frac{a^{2}}{x^{2}} \right) \ dx &= \int_{0}^{\infty} \cos \left(\frac{a^{2}}{x^{2}} - b^{2} x^{2} \right) \ dx \\ &= \frac{1}{2b} \sqrt{\frac{\pi}{2}} e^{-2 ab} . \end{align}$$

$\endgroup$
3
$\begingroup$

I confirm that, before any simplification, $$I=\int_0^{\infty} e^{i \left(\frac{a^2}{x^2}-b^2 x^2\right)} \, dx=\frac{\sqrt{\pi } e^{-\frac{2 \sqrt{i b^2}}{\sqrt{\frac{i}{a^2}}}}}{2 \sqrt{i b^2}}$$ under the conditions that $\Im\left(b^2\right)<0\land \Im\left(a^2\right)>0$.

After simplifications $$J=\int_0^{\infty} \cos\left(\frac{a^2}{x^2}-b^2x^2\right)\,dx=\Re\left(\frac{1}{2b}\sqrt{\frac{\pi}{i}}e^{-2ab}\right)=\frac{1}{2b}\sqrt{\frac{\pi}{2}}e^{-2ab}$$ which is your answer with a minus sign in the exponent.

I performed numerical checks and this is correct.

$\endgroup$
  • $\begingroup$ I am really sorry, I forgot to mention the constraints on $a$ and $b$. For the given problem, $a,b>0$. Does your result holds true for this case? Also, in my attempt, I realised that I cannot write $-i$ as $i^3$. I can write it as $i^7$ and this will give me the correct result. However, how do I decide between $i^3$ and $i^7$? Thank you! $\endgroup$ – Pranav Arora Jul 13 '14 at 4:31
  • $\begingroup$ I do not see why you used $i$ and $i^3$. I just used $i$ and everything came. $\endgroup$ – Claude Leibovici Jul 13 '14 at 4:34
  • $\begingroup$ No no, I didn't do that. Let me explain more clearly. In the exponent of $e$, I have: $$i\left(\frac{a^2}{x^2}-b^2x^2\right)=-\left(ib^2x^2-i\frac{a^2}{x^2}\right)$$ I can write $-i$ as $i^3$ or $i^7$. If I write $-i$ as $i^3$, I get (in the exponent of $e$) $-2\cdot i^{1/2}b\cdot i^{3/2} a=2ab$ and if I use $i^7$, I get: $-2\cdot i^{1/2}b \cdot i^{7/2}a=-2ab$. I do not understand the reason behind these different results. $\endgroup$ – Pranav Arora Jul 13 '14 at 4:40
  • $\begingroup$ Ah, the problem is with the nature of complex exponentiation. It is multivalued due to the fact that $\exp$ is not injective on the entire complex plane. Thus when you take the square root of $i^7$ and $i^3$ as you would for real numbers, you get different answers. (The reason being you assume $\sqrt{1} = \sqrt{i^4} = i^2 = -1$, which, while certainly being a root of $1$, isn't necessarily the one you want). $\endgroup$ – Joshua Mundinger Jul 14 '14 at 15:42

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.