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As far as I know that the Gaussian Integral $$\int_{- \infty}^{\infty} e^{-ax^2} = \frac{\pi}{a}$$ And using the above integration the Author D. Griffiths computed the integration (7.3) and (7.5). But I find contradiction in the equation (7.5). I used the same formula but got the result

$$-\frac{2 \hbar^2 b^2}{m}$$ But they got different. If their computation was a typo they won't use again and again, but they used it and got the final result correct. Am I doing wrong somewhere in the equation (7.5)? enter image description here

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First note that

$$\int_{-\infty}^\infty e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}} \tag 1$$

Next, by differentiating under the integral, we find that

$$-\frac{d}{da}\int_{-\infty}^\infty e^{-ax^2}\,dx=\int_{-\infty}^\infty x^2e^{-ax^2}\,dx=\sqrt{\frac{\pi}{4a^3}} \tag 2$$

Using $(1)$ and $(2)$ reveals that

$$\begin{align} \int_{-\infty}^\infty e^{-bx^2}\,\frac{d^2}{dx^2}\left(e^{-bx^2}\right)\,dx&=\int_{-\infty}^\infty e^{-2bx^2}(4b^2x^2-2b)\,dx\\\\ &=4b^2\int_{-\infty}^\infty x^2e^{-2bx^2}\,dx-2b\int_{-\infty}^\infty e^{-2bx^2}\,dx\\\\ &=4b^2\sqrt{\frac{\pi}{32b^3}}-2b\sqrt{\frac{\pi}{2b}}\\\\ &=-\sqrt{\frac{\pi b}{2}} \end{align}$$

which after multiplication by $-\frac{\bar h^2}{2m}|A|^2$ gives the reported result.

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