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[Calculus] Bypassing Cyclic Integration by Parts

2021-09-19 12:36 作者:AoiSTZ23 0人讀過 | 我要投稿

By: Tao Steven Zheng (鄭濤)

【Problem】

Use the integral

$%5Cint%20%7Be%7D%5E%7B(a%2Bib)x%7D%20dx%20%3D%20%5Cfrac%7B%7Be%7D%5E%7B(a%2Bib)x%7D%7D%7Ba%2Bib%7D$

to compute $%5Cint%20%7Be%7D%5E%7Bax%7D%5Csin(bx)%20dx$ and $%5Cint%20%7Be%7D%5E%7Bax%7D%5Ccos(bx)%20dx%20$ simultaneously.

【Solution】

Since $%20%7Be%7D%5E%7B(a%2Bib)x%7D%20%3D%20%7Be%7D%5E%7Bax%7D%20%5B%5Ccos(bx)%20%2B%20%20i%5Csin(bx)%5D$ ,

$%5Cint%20%7Be%7D%5E%7Bax%7D%20%5B%5Ccos(bx)%20%2B%20%20i%5Csin(bx)%5D%20dx%20%3D%20%5Cfrac%7B%7Be%7D%5E%7B(a%2Bib)x%7D%7D%7Ba%2Bib%7D$

$%5Cint%20e%5E%7Bax%7D%5Ccos(bx)dx%20%2B%20%20i%20%5Cint%20e%5E%7Bax%7D%5Csin(bx)%20dx%20%3D%20%5Cfrac%7B%7Be%7D%5E%7Bax%7D%20%5B%5Ccos(bx)%20%2B%20%20i%5Csin(bx)%5D%7D%7Ba%2Bib%7D$

Multiply the numerator and denominator by the complex conjugate on the right hand side.

$%5Cbegin%7Balign%7D%0A%5Cint%20e%5E%7Bax%7D%5Ccos(bx)dx%20%2B%20%20i%20%5Cint%20e%5E%7Bax%7D%5Csin(bx)%20dx%20%26%3D%20%5Cfrac%7B%7Be%7D%5E%7Bax%7D%20%5B%5Ccos(bx)%20%2B%20%20i%5Csin(bx)%5D(a-ib)%7D%7B(a%2Bib)(a-ib)%7D%5C%5C%0A%26%3D%20%5Cfrac%7B%7Be%7D%5E%7Bax%7D%7D%7Ba%5E2%20%2B%20b%5E2%7D%20%5Ba%5Ccos(bx)%20%2B%20%20ia%5Csin(bx)-ib%5Ccos(bx)%20%2B%20b%5Csin(bx)%5D%5C%5C%0A%5Cend%7Balign%7D$

By matching real terms and complex terms, and tacking on the arbitrary constant from integrating, we can conclude that

$%5Cint%20%7Be%7D%5E%7Bax%7D%5Csin(bx)%20dx%20%3D%20%5Cfrac%7B%7Be%7D%5E%7Bax%7D%20%5Ba%5Csin(bx)%20-%20%20b%5Ccos(bx)%5D%7D%7Ba%5E2%20%2B%20b%5E2%7D%0A%20%2B%20C$

and
$%5Cint%20%7Be%7D%5E%7Bax%7D%5Ccos(bx)%20dx%20%3D%20%5Cfrac%7B%7Be%7D%5E%7Bax%7D%20%5Ba%5Ccos(bx)%20%2B%20%20b%5Csin(bx)%5D%7D%7Ba%5E2%20%2B%20b%5E2%7D%0A%20%2B%20C%20$