Existence of closed form solutions on a definite integral.

Asked May 03 '13 at 16:00

Active May 03 '13 at 16:26

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I need to find the value of the given expression in a closed form

$$\int_0^{\frac{\pi}{2}} \dfrac{\sin x}{ x} dx$$

My effort:

I know that its a standard definite integral called $Si(x)$

$\sin x= x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+ \dots$

The integral of the expression becomes

$x-\dfrac{x^3}{3\cdot3!}+\dfrac{x^5}{5\cdot5!}-\dfrac{x^7}{7\cdot7!}+ \dots$

Now I just need to put limits. The answer what wolframalpha says is $Si(\dfrac\pi 2)\approx 1.37076$

I just want to know whether there's a closed form or a pretty answer to this

$\lim_{x \to \frac{\pi}{2}}(x-\dfrac{x^3}{3\cdot3!}+\dfrac{x^5}{5\cdot5!}-\dfrac{x^7}{7\cdot7!}+ \dots$) ? Something like $\dfrac{e}{2}$?

edited May 03 '13 at 16:26

asked May 03 '13 at 16:00

Inceptio

may be you can try wallis formula – Tao May 03 '13 at 16:06
Well, great hint. But I gotta find the limit of the integral and the Wallis product surely helps if it was just to find the $\lim_{x \to \frac{\pi}{2}} (\dfrac{\sin x}{x})$. – Inceptio May 03 '13 at 16:12
1

I would be very surprised if there were a closed form answer to this one. – Harald Hanche-Olsen May 03 '13 at 16:14
@HaraldHanche-Olsen: I doubt whether there is any closed form solutions. – Inceptio May 03 '13 at 16:25
You can have a solution in terms of the sine integral $Si(\pi/2)$. – Mhenni Benghorbal May 04 '13 at 03:41

0 Answers0