This question was inspired by this question:

Evaluating the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?

Well, can anyone prove this without using Residue theory. I actually thought of doing this: $$\int_{0}^{\infty} \frac{\sin x}{x} \, dx = \lim_{t \to \infty} \int_{0}^{t} \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \, dt$$ but I don't see how $\pi$ comes here, since we need the answer to be equal to $\frac{\pi}{2}$.

Answers were given to the stated question -- how to prove without using Residue theory. Yet the quote suggests an obvious follow-up question: can you prove the integral from the Taylor series expansion directly, somehow?

  • $\begingroup$ The exact same question can be asked about the Gaussian integral, and the answer in both cases is a no. $\endgroup$
    – Lucian
    Sep 17 '15 at 8:23
  • $\begingroup$ @Lucian: Interesting. Can you elaborate? (that is, post an answer detailing why) $\endgroup$ Sep 17 '15 at 8:30
  • $\begingroup$ There are obvious convergence issues since the integral of $\frac{x^{2k}}{C}$ over $\mathbb{R}^+$ is $+\infty$ no matter how big the constant $C>0$ is. $\endgroup$ Sep 17 '15 at 14:28
  • $\begingroup$ This may interest you: web.williams.edu/Mathematics/lg5/Feynman.pdf $\endgroup$ Sep 17 '15 at 14:55

HINT: Notice, $$\lim_{t\to \infty}\int_{0}^{t}\frac{1}{t}\left(t-\frac{t^3}{3!}+\frac{t^5}{5!}+\dots \right)dt$$ $$=\lim_{t\to \infty}\int_{0}^{t}\frac{1}{t}\left(t-\frac{t^3}{3!}+\frac{t^5}{5!}+\dots \right)dt$$ $$=\lim_{t\to \infty}\int_{0}^{t}\left(1-\frac{t^2}{3!}+\frac{t^4}{5!}+\dots \right)dt$$ $$=\lim_{t\to \infty}\left(t-\frac{t^3}{3\cdot3!}+\frac{t^5}{5\cdot 5!}+\dots \right)_{0}^{t}$$ $$=\lim_{t\to \infty}\left(t-\frac{t^3}{3\cdot3!}+\frac{t^5}{5\cdot 5!}+\dots \right)$$

  • $\begingroup$ How do you conclude from this? $\endgroup$
    – mickep
    Sep 19 '15 at 18:26

This post does not derive the result using only Taylor series, but it only uses the fundamental property of complex numbers, that $z = a+bi$.

Euler derived a awesome formula

$$\frac{\Gamma (s)}{n|p|^s}\sin{(\alpha s)}=\int_0^{\infty}u^{ns-1}\exp{(-\Re(p)u^n)}\sin{(\Im(p)u^n)}du$$

Where $tan(\alpha)=\frac{\Im(p)}{\Re(p)}$ Here is an informal proof of the formula https://goo.gl/JKfEO8.

Now use $n=1$, $p=0+i$, $s=0$, $\alpha = \frac{\pi}{2}$. You will also need Eulers Reflection Formula for the Gamma function $$\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin{(\pi s)}}$$

If you have trouble solving the integral using the formula you can also watch this https://goo.gl/2WPRQH.

  • $\begingroup$ This is very nice indeed! but can we use it to answer the question? $\endgroup$
    – Math-fun
    Sep 17 '15 at 8:03
  • $\begingroup$ Yes you can, follow the link. You can also use this formula to calculate the Fresnel Integrals. $\endgroup$
    – MrYouMath
    Sep 17 '15 at 8:05
  • $\begingroup$ I just did so :-) but this technique does not help with "direct" calculation of $\lim_{a \to \infty} \int_{0}^{a} \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \, dt$ $\endgroup$
    – Math-fun
    Sep 17 '15 at 8:10
  • $\begingroup$ I know it is not a direct way :). And i am also looking forward for someone to directly calculate it :D. Acutally the problem is the same as showing that $$\lim_{a\to \infty}\sum_{n=1}^{\infty}(-1)^{n+1}\frac{a^{2n-1}}{(2n-1)(2n-1)!}=\frac{\pi}{2}$$ $\endgroup$
    – MrYouMath
    Sep 17 '15 at 8:19
  • $\begingroup$ I think if there is a way to show it using the taylor series you also will need to use the Euler Product Formula for sine. $\endgroup$
    – MrYouMath
    Sep 17 '15 at 8:25

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