5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – The_Sympathizer 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. $\endgroup$ – Jack D'Aurizio Sep 17 '15 at 14:28
  • $\begingroup$ This may interest you: web.williams.edu/Mathematics/lg5/Feynman.pdf $\endgroup$ – Salihcyilmaz Sep 17 '15 at 14:55
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ How do you conclude from this? $\endgroup$ – mickep Sep 19 '15 at 18:26
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.