[Calculus] Curious Integral of Johann Bernoulli

2022-06-10 07:36 作者:AoiSTZ23 0人讀過(guò) | 我要投稿

By: Tao Steven Zheng (鄭濤)

【Problem】?

In 1697, the Swiss mathematician Johann Bernoulli?(1667 - 1748) discovered a very interesting result of integral calculus:

$%5Cint_%7B0%7D%5E%7B1%7D%20%5Cfrac%7B1%7D%7Bx%5Ex%7D%20dx%20%3D%20%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B1%7D%7Bn%5En%7D.%20$

Prove this interesting result!

Hints

(1) Begin by using the transformation $%20%5Cfrac%7B1%7D%7Bx%5Ex%7D%20%3D%20%7Be%7D%5E%7B-x%20%5Cln%20x%7D$ .

(2) Remember the Taylor series expansion $e%5Ex%20%3D%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bx%5En%7D%7Bn!%7D$ .

【Solution】?

This integral is an improper integral with the indeterminate form $0%5E0$ . It is well known that $%5Clim_%7Bx%20%5Crightarrow%200%7D%20x%5Ex%20%3D%201$ and knowing this fact enables us to show that

$%20%5Cint_%7B0%7D%5E%7B1%7D%20%5Cfrac%7B1%7D%7Bx%5Ex%7D%20dx%20%3D%20%5Cint_%7B0%7D%5E%7B1%7D%20%7Be%7D%5E%7B-x%20%5Cln%20x%7D%20dx.$

Applying the Taylor series $e%5Ex%20%3D%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7Bx%5En%7D%7Bn!%7D$ gives

$%5Cbegin%7Balign%7D%20%5Cint_%7B0%7D%5E%7B1%7D%20%5Cfrac%7B1%7D%7Bx%5Ex%7D%20dx%20%26%3D%20%5Cint_%7B0%7D%5E%7B1%7D%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B%7B(-x%20%5Cln%20x)%7D%5E%7Bn%7D%7D%7Bn!%7D%20dx%20%5C%5C%0A%0A%26%3D%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B1%7D%7Bn!%7D%20%5Cint_%7B0%7D%5E%7B1%7D%20%7B(-x%20%5Cln%20x)%7D%5E%7Bn%7D%20dx%20%5C%5C%0A%0A%26%3D%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B%7B(-1)%7D%5E%7Bn%7D%7D%7Bn!%7D%20%5Cint_%7B0%7D%5E%7B1%7D%20x%5En%20%7B%5Cln%7D%5E%7Bn%7D%20x%20%5C%3Bdx%20%5C%5C%0A%0A%5Cend%7Balign%7D%20$

Now use the technique of integration by parts on the integral $%5Cint_%7B0%7D%5E%7B1%7D%20x%5En%20%7B%5Cln%7D%5E%7Bn%7D%20x%20%5C%3Bdx$ . Let $u%20%3D%20%7B%5Cln%7D%5E%7Bn%7D%20x$ , $du%20%3D%20%5Cfrac%7Bn%7D%7Bx%7D%20%7B%5Cln%7D%5E%7Bn-1%7D%20x%20%5C%3Bdx$ , $dv%20%3D%20x%5En%20dx%20$ , and $v%20%3D%20%5Cfrac%7B%7Bx%7D%5E%7Bn%2B1%7D%7D%7Bn%2B1%7D$ .?

Subsequently,

$%20%5Cbegin%7Balign%7D%20%5Cint_%7B0%7D%5E%7B1%7D%20x%5En%20%7B%5Cln%7D%5E%7Bn%7D%20x%20%5C%3Bdx%20%26%3D%20%5Cleft(%5Cfrac%7B%7Bx%7D%5E%7Bn%2B1%7D%7D%7Bn%2B1%7D%7B%5Cln%7D%5E%7Bn%7D%20x%20%5Cright)_%7B0%7D%5E%7B1%7D%20-%20%5Cfrac%7Br%7D%7Br%2B1%7D%5Cint_%7B0%7D%5E%7B1%7D%20%5Cfrac%7B%7Bx%7D%5E%7Bn%2B1%7D%7D%7Bx%7D%20%7B%5Cln%7D%5E%7Bn-1%7D%20x%20%5C%3Bdx%20%5C%5C%0A%0A%26%3D%20-%5Cfrac%7Bn%7D%7Bn%2B1%7D%20%5Cint_%7B0%7D%5E%7B1%7D%20%7Bx%7D%5E%7Bn%7D%20%7B%5Cln%7D%5E%7Bn-1%7D%20x%20%5C%3Bdx%20%5C%5C%0A%0A%5Cend%7Balign%7D%20$

Continuing the integration by parts scheme to the integral $%20%5Cint_%7B0%7D%5E%7B1%7D%20%7Bx%7D%5E%7Bn%7D%20%7B%5Cln%7D%5E%7Bn-1%7D%20x%20%5C%3Bdx$ gives the pattern:

$%5Cbegin%7Balign%7D%20%0A%0A%5Cint_%7B0%7D%5E%7B1%7D%20%7Bx%7D%5E%7Bn%7D%20%7B%5Cln%7D%5E%7Bn-1%7D%20x%20%5C%3Bdx%20%26%3D%20-%5Cfrac%7Bn-1%7D%7Bn%2B1%7D%20%5Cint_%7B0%7D%5E%7B1%7D%20%7Bx%7D%5E%7Bn%7D%20%7B%5Cln%7D%5E%7Bn-2%7D%20x%20%5C%3Bdx%20%5C%5C%0A%0A%5Cint_%7B0%7D%5E%7B1%7D%20%7Bx%7D%5E%7Bn%7D%20%7B%5Cln%7D%5E%7Bn-2%7D%20x%20%5C%3Bdx%20%26%3D%20-%5Cfrac%7Bn-2%7D%7Bn%2B1%7D%5Cint_%7B0%7D%5E%7B1%7D%20%7Bx%7D%5E%7Bn%7D%20%7B%5Cln%7D%5E%7Bn-3%7D%20x%20%5C%3Bdx%2C%20%5C%3Betc.%5C%5C%0A%0A%5Cend%7Balign%7D$

Thus,

Consequently,

$%5Cbegin%7Balign%7D%0A%0A%5Cint_%7B0%7D%5E%7B1%7D%20x%5En%20%7B%5Cln%7D%5E%7Bn%7D%20x%20%5C%3Bdx%20%26%3D%20%5Cfrac%7B%7B(-1)%7D%5E%7Bn%7Dn!%7D%7B%7B(n%2B1)%7D%5E%7Bn%7D%7D%20%5Cint_%7B0%7D%5E%7B1%7D%20%7Bx%7D%5E%7Bn%7D%20dx%20%5C%5C%0A%0A%26%3D%20%5Cfrac%7B%7B(-1)%7D%5E%7Bn%7Dn!%7D%7B%7B(n%2B1)%7D%5E%7Bn%2B1%7D%7D%20%0A%0A%5Cend%7Balign%7D$

Therefore,

$%5Cint_%7B0%7D%5E%7B1%7D%20%5Cfrac%7B1%7D%7Bx%5Ex%7D%20dx%20%3D%20%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B1%7D%7Bn%5En%7D.%20$

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[Calculus] Curious Integral of Johann Bernoulli

By: Tao Steven Zheng (鄭濤)

【Problem】?

【Solution】?