ai giúp tui cái coi
#1
Đã gửi 08-07-2008 - 00:12
#2
Đã gửi 12-07-2008 - 00:49
tính $\int\dfrac{dx}{x^{8}+1}$
Bạn có thể tham khảo lời giài cho trường hợp tổng quát ở đây:
http://mathforum.org...view/53784.html
Kq: http://integrals.wol...mp;random=false
#3
Đã gửi 13-07-2008 - 08:07
#4
Đã gửi 13-07-2008 - 13:04
Integrating 1/(1+x^n)
Date: 11/14/2001 at 17:09:51
From: Varghese Devassy
Subject: Integration
Hi,
I was wondering whether there are methods to integrate functions of
the form 1/(1+x^n). I have solved 1/(1+x^4) and am wondering about
solutions for higher powers of x.
Thank you,
Varghese Devassy
Date: 11/16/2001 at 04:22:36
From: Doctor Pete
Subject: Re: Integration
Hi,
Let F[x,n]= x^n + 1. Then 1/F has an "elementary" antiderivative for
rational values of n. However, let's consider the case where n is a
positive integer. The idea is to express F as a product of linear
factors; specifically,
F[x,n] = (x - r[1])(x - r[2])...(x - r[n]),
where r[1], r[2], ..., r[n] are the roots of F. Then we use partial
fraction decomposition to express 1/F as the sum of terms of the form
A[k]/(x - r[k]),
for suitable coefficients A[k].
With this in mind, the roots of F are
{Exp[I*Pi/n], ..., Exp[(2k-1)I*Pi/n], ..., Exp[(2n-1)I*Pi/n],
where k takes on integer values from 1 to n. For the moment, never
mind that these are complex numbers. Then we have
r[k] = Exp[(2k-1)I*Pi/n].
The partial fraction decomposition step is tricky. The general term
of the decomposition is A[k]/(x - r[k]), for some coefficient A[k].
To find this coefficient, we notice that cross-multiplication and
substitution of the value x = r[k] gives
A[k](r[k] - r[1])...(r[k] - r[k-1])(r[k] - r[k+1])...(r[k] -
r[n]) = 1,
where the product of linear factors on the left-hand side excludes the
factor (r[k] - r[k]), which is zero. This product happens to be
-n*Exp[-(2k-1)I*Pi/n],
so A[k] = (-1/n)Exp[(2k-1)I*Pi/n] = -r[k]/n. (The proof of this I
leave to you.)
Now what? So far, we have expressed 1/F as the sum of linear factors
of the form
(-r[k]/n)/(x - r[k]), r[k] = Exp[(2k-1)I*Pi/n],
for k = 1 to k = n. Motivated by the idea that complex conjugate
pairs add and multiply to real numbers, let us call s[k] the complex
conjugate to r[k]. Note that
{s[1], s[2], ..., s[n]} = {r[1], r[2], ..., r[n]};
i.e., the set of s[k]'s is the same as the set of r[k]'s, because the
complex roots of F come in conjugate pairs. So if we have the pair of
terms
(-r[k]/n)/(x - r[k]) + (-s[k]/n)/(x - s[k]),
cross-multiplying gives
((-r[k]/n)(x - s[k]) + (-s[k]/n)(x - r[k]))/((x - r[k])(x -
s[k])).
Now, note that
r[k] = Exp[(2k-1)I*Pi/n] = Cos[(2k-1)Pi/n] + I*Sin[(2k-1)Pi/n],
q[k] = Exp[(1-2k)I*Pi/n] = Cos[(2k-1)Pi/n] - I*Sin[(2k-1)Pi/n],
so in particular we have
r[k] + q[k] = 2 Cos[(2k-1)Pi/n],
r[k]q[k] = 1.
Therefore, the sum of the conjugate pair terms is
(-1/n)(r[k]x - 1 + q[k]x - 1)/(x^2 - (r[k]+q[k])x + 1)
= (2/n)(1 - Cos[(2k-1)Pi/n]x)/(x^2 - 2 Cos[(2k-1)Pi/n]x + 1).
Note that this expression is real. So it follows that
2/F[x,n] = Sum[(2/n)(1-Cos[(2k-1)Pi/n]x)/(x^2 - 2Cos[(2k-1)Pi/n]x
+ 1)]
where the sum is taken over k = 1 to k = n. Now we can see where we're
going, for from here it is a relatively simple matter to integrate
this expression term by term. Let
C[k] = Cos[(2k-1)Pi/n],
S[k] = Sin[(2k-1)Pi/n],
so the k(th) term of the integrand is
(2/n)(1-C[k]x)/(x^2 - 2C[k]x + 1).
Before we integrate, however, note that if C[k] = -1 (which happens
when n is odd), the integrand simplifies to (2/n)/(x+1), which
integrates to
(-2/n)Log[x+1].
Otherwise, we separate the integrand into two parts, giving
-(C[k]/n)(2x - 2C[k])/(x^2-2C[k]x+1) + (2S[k]^2/n)/(x^2-2C[k]x+1)
The first term is easily integrated with the substitution
u = x^2 - 2C[k]x + 1, du = (2x - 2C[k]) dx,
giving
-(C[k]/n)Log[x^2 - 2C[k]x + 1].
The second term is a bit harder. Completing the square in the
denominator, we obtain
(2S[k]^2/n)/(x^2 - 2C[k]x + C[k]^2 + S[k]^2)
= (2S[k]^2/n)/((x - C[k])^2 + S[k]^2).
With the substitution u = x - C[k], we obtain
(2S[k]^2/n)/(u^2 + S[k]^2),
which directly integrates to
(2S[k]/n) ArcTan[(x-C[k])/S[k]].
Therefore, the integral of 1/F is half the sum of the above terms from
k = 1 to k = n, which is
(S[k]/n)ArcTan[(x-C[k])/S[k]] - (C[k]/(2n))Log[x^2-2C[k]x+1],
whenever S[k] is nonzero (which implies C[k] is not -1), and
(1/n)Log[x+1]
when S[k] = 0 and C[k] = -1.
Whew! I hope I didn't make a typing error, and I hope you understood
my line of reasoning. Notice that most of the work was in expressing
the integrand in a form that was readily integrable; only in the very
last step did I actually integrate the function. Also note that I left
a little fact unproven, and it might be a good exercise to try to
prove it.
Bài viết đã được chỉnh sửa nội dung bởi Zerocool: 13-07-2008 - 13:06
1 người đang xem chủ đề
0 thành viên, 1 khách, 0 thành viên ẩn danh