Take the integral $$ \int_{-\infty}^{\infty}e^{-\lambda x^2}\frac{d^2}{dx^2}(e^{-\lambda x^2})dx = \int_{-\infty}^{\infty}e^{-\lambda x^2}\frac{d}{dx}(-2 \lambda x e^{-\lambda x^2})dx = \int_{-\infty}^{\infty}(-2 \lambda + 4 \lambda^2 x^2 )e^{-2\lambda x^2}dx = -\lambda\frac{\sqrt{\pi}}{(2\lambda)^{3/2}} + 2 \lambda^2\frac{\sqrt{\pi}}{(2\lambda)^{3/2}} $$

But looking at the first expression and applying partial integration we get: $$ \int_{-\infty}^{\infty}e^{-\lambda x^2}\frac{d^2}{dx^2}(e^{-\lambda x^2})dx = [e^{-\lambda x^2}\frac{d}{dx}(e^{-\lambda x^2})]_{-\infty}^{\infty} - \int_{-\infty}^{\infty}(-2) \lambda x e^{-\lambda x^2}\frac{d}{dx}(e^{-\lambda x^2})dx = $$

$$ -\int_{-\infty}^{\infty}4 \lambda^2 x^2 e^{-2\lambda x^2}dx = - 2 \lambda^2\frac{\sqrt{\pi}}{(2\lambda)^{3/2}} $$

Where by taking limits:

$$ [e^{-\lambda x^2}\frac{d}{dx}(e^{-\lambda x^2})]_{-\infty}^{\infty} = 0 $$

This doesn't seem correct because $$ - 2 \lambda^2\frac{\sqrt{\pi}}{(2\lambda)^{3/2}} \neq -\lambda\frac{\sqrt{\pi}}{(2\lambda)^{3/2}} + 2 \lambda^2\frac{\sqrt{\pi}}{(2\lambda)^{3/2}} $$

equivalent to

$$ \lambda \neq 4\lambda^2 $$

What is going on here?

  • 1
    $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Shaun May 24 at 15:50
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    $\begingroup$ Definite integrals of two different functions may perfectly coincide, while the functions are different. Say $\int_0^1 2x dx = \int_0^1 1 dx$. Equality of the definite integrals does not imply equality of the functions. $\endgroup$ – guest May 24 at 15:53
  • $\begingroup$ Why you think it's incorrect as both the integrals are 0 $\endgroup$ – Aditya Dwivedi May 24 at 15:54
  • $\begingroup$ Thank you for your comments, I realise the integrals could be equal. However, I calculated the integrals and still have inequality, very annoying. See edited question. $\endgroup$ – Oskar Söderberg May 24 at 16:24

$$ \int_{-\infty}^{\infty}(-2 \lambda)e^{-2\lambda x^2}dx = -\sqrt{2\pi\lambda} \neq -\lambda\frac{\sqrt{\pi}}{(2\lambda)^{3/2}}. $$

Taken together with the fact that $$ 2 \lambda^2\frac{\sqrt{\pi}}{(2\lambda)^{3/2}} = \frac12 \sqrt{2\pi\lambda}, $$ the inconsistency in your results is resolved.

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Attaching some fictitious "physical dimension" to the objects helps in this case: it allows you to find a possible mistake.

Say that $x \sim L$ is a "length". The argument of the exponential must be a pure number (namely $\lambda x^2 \sim L^0$), so that $\lambda \sim L^{-2}$. Clearly, $dx \sim L$ and $d/dx^2 \sim L^{-2}$.

Therefore, the result of the integral (call it $I$) must be

$$I \sim L^{-1}.$$

This is not consistent with what you have in your first equation. You probably made a mistake in the calculation: you cannot add terms with different physical dimensions (i.e. $L+L^{-1}$ does not make sense).

You should obtain

$$I = -\sqrt{\pi\lambda /2} \sim L^{-1} \qquad \qquad Re(\lambda)>0,$$ which is what you get with your second approach (the integration by parts).

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