Integral estimate

(FB/Math/BM – Question raised by Sachin Modi)

Is the integral  ∫ [π,3π] |sin x + cos x|/x dx less than or greater than 1 ?

_________________________________________________

The integral equals

∫ [-π/4,π/2] (√2·cos x)/(x + 5π/4) dx
+ ∫ [-π/2,π/2] (√2·cos x)/(x + 9π/4) dx
+ ∫ [0,π/4] (√2·sin x)/(x + 11π/4) dx

Taylor expansion shows that 1/(x + 5π/4) > (4/(5π))·(1 – 4x/(5π)) etc. and the integrands are non-negative in the respective intervals. Thus,

∫ [-π/4,π/2] (√2·cos x)/(x + 5π/4) dx
> (4/(5π))·∫ [-π/4,π/2] (√2·cos x)(1 – 4x/(5π)) dx =

= (12·√2/(25·π) + 24/(25·π) + 16/(25·π²))

and similarly for the other two integrals, and summing up,

∫ [π,3π](|sinx +cosx|/x)dx
> 4288·√2/(2475·π) + 1904/(3025·π) + 1536/(3025·π²) > 1, so
—————————-
∫ [π,3π](|sinx +cosx|/x)dx > 1.
—————————-
4288·√2/(2475·π) + 1904/(3025·π) + 1536/(3025·π²) ≈ 1.031709173 which I think/hope is sufficiently far from 1 to rely on the result.

_______________________

It could also be mentioned that
4288·√2/(2475·π) + 1904/(3025·π) + 1536/(3025·π²) > 16857176/16471125 > 1, using π < 22/7 och √2 > 7/5.

For convenience I used Derive (of course) though there is no great problem doing all by hand; the i…ntegrand are not more difficult than x·cos x.

Also Taylor expansion is not really nedded; I just use 1/(1 + x) > 1 – x.

Annonser

Kommentera

Fyll i dina uppgifter nedan eller klicka på en ikon för att logga in:

WordPress.com Logo

Du kommenterar med ditt WordPress.com-konto. Logga ut /  Ändra )

Google+-foto

Du kommenterar med ditt Google+-konto. Logga ut /  Ändra )

Twitter-bild

Du kommenterar med ditt Twitter-konto. Logga ut /  Ändra )

Facebook-foto

Du kommenterar med ditt Facebook-konto. Logga ut /  Ändra )

Ansluter till %s

%d bloggare gillar detta: