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The definite integral $\int_0^\infty e^{-x^2}\mathrm{d}x=\frac{\sqrt{\pi}}{2}$ is very typical. We have:

$$1-x^2 \leqslant e^{-x^2}(0 \leqslant x \leqslant 1), \quad e^{-x^2} \leqslant \frac{1}{1+x^2} \quad (x \geqslant 0)$$

Besides, the result:

$$\lim_\limits{n \rightarrow \infty} \sqrt{n} \int_{0}^{1} (1 - x^2)^n \mathrm{d}x=\frac{\sqrt{\pi}}{2}$$

can be got by Wallis formula. And my proof is as follows:

For the two inequalities above, we have:

$$\int_{0}^{1} (1-x^2)^n \mathrm{d}x < \int_{0}^{1} e^{-nx^2} \mathrm{d}x < \int_{0}^{1} \frac{\mathrm{d}x}{(1+x^2)^n} < \int_{0}^{\infty} \frac{\mathrm{d}x}{(1+x^2)^n}$$

so

$$\sqrt{n}\int_{0}^{1} (1-x^2)^n \mathrm{d}x < \int_{0}^{\sqrt{n}} e^{-t^2} \mathrm{d}t < \sqrt{n}\int_{0}^{\infty} \frac{\mathrm{d}x}{(1+x^2)^n}$$

Then I let $n \rightarrow \infty$ and want to utilize the sandwish theorem for proof. But I don't know how to calculate the right definite integral:

$$\sqrt{n}\int_{0}^{\infty} \frac{\mathrm{d}x}{(1+x^2)^n}$$

Can you help me?

ZixuanLing
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  • Wolfram Alpha suggests $\frac{\sqrt{\pi}}{2} \frac{\sqrt{n}\Gamma(n-1)}{\Gamma(n)}$ which will also have a limit of $\frac{\sqrt{\pi}}{2}$ – Henry Mar 19 '23 at 12:10
  • Initial guess would be to substitute $x=\tan t.$ – Thomas Andrews Mar 19 '23 at 12:13
  • $$\sqrt{n} \int_{0}^{1} \frac{\mathrm{d}x}{(1+x^2)^n} = \sqrt{n} \int_{0}^{\frac{\pi}{4}} \frac{\sec^2 t}{(1+\tan^2 t)^n} \mathrm{d}t$$ $$\sqrt{n} \int_{0}^{\frac{\pi}{4}} \frac{1+\tan^2 t}{(1+\tan^2 t)^n} \mathrm{d}t = \sqrt{n} \int_{0}^{\frac{\pi}{4}} (1+\tan^2 t)^{1-n} \mathrm{d}t$$ $$B(p, q) = \int_{0}^{\frac{\pi}{4}} (\tan t)^{p-1}(1+\tan^2 t)^{-p-q} \mathrm{d}t = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} = \sqrt{n} \cdot \frac{\Gamma(2-n)\Gamma(n)}{\Gamma(2)}$$ – rumathe Mar 19 '23 at 12:26
  • $$\lim_{n \to \infty} \frac{\sqrt{\pi}}{2} \frac{\sqrt{n}\Gamma(n-1)}{\Gamma(n)} = \frac{\sqrt{\pi}}{2} \lim_{n \to \infty} \frac{\sqrt{n}\Gamma(n-1)}{\Gamma(n)}$$ $$\lim_{n \to \infty} \frac{\sqrt{\pi}}{2} \frac{\sqrt{n}\Gamma(n-1)}{n\Gamma(n-1)} = \frac{\sqrt{\pi}}{2} \lim_{n \to \infty} \frac{\sqrt{n}}{n}$$ $$\frac{\sqrt{\pi}}{2} \lim_{n \to \infty} \frac{1}{\sqrt{n}} = \frac{\sqrt{\pi}}{2}$$ – rumathe Mar 19 '23 at 12:26
  • Look here – Sine of the Time Mar 19 '23 at 12:46
  • See the seventh & eighth proofs here. – J.G. Mar 19 '23 at 12:47

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