An Useful Integral Formula

Oct 2010
492
14
Changchun, China
We do meet the integrals such as \(\displaystyle \int x^n\cos{(ax+b)}dx\)or \(\displaystyle \int x^n\sin{(ax+b)}dx\)frequently in

calculus area., where n is a positive integer.

Actually, the two above are just the same, because \(\displaystyle \sin{(ax+b)}=\cos{(ax+b-\frac{\pi}{2})}=\cos{(ax+c)}\), where

\(\displaystyle c=b-\frac{\pi}{2}\).

So, \(\displaystyle I_n=\int x^n\cos{(ax+b)}dx\) is the only one I want to discuss here.

Firstly I give the formula as below:

\(\displaystyle I_n=\int x^n\cos{(ax+b)}dx =\sum_{m=0}^{n}{\frac{\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\sin{(ax+b+\frac{m\pi}{2})}\),

where \(\displaystyle \Gamma(.+1)=(.)!\)

Proof:

It is easy to show that, the integral:

\(\displaystyle \int x^n\e{(ax+b)}dx\)

\(\displaystyle =\sum_{m=0}^{n}{\frac{(-1)^{m}\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\e (ax+b)\)

By using the Euler's formula, \(\displaystyle \cos t=\frac{\e (-it)+\e (it)}{2}\)(\(\displaystyle i\) denotes the imaginary unit):

\(\displaystyle I_n=\int x^n\cos{(ax+b)}dx\)

\(\displaystyle =\frac{1}{2}\int x^n(\e{[-i(ax+b)]}+\e{[i(ax+b)]})dx\)

\(\displaystyle =\frac{1}{2}\{\sum_{m=0}^{n}{\frac{(-1)^{m}\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)(-ia)^{m+1}}\e {[-i(ax+b)]}+\sum_{m=0}^{n}{\frac{(-1)^{m}\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)(ia)^{m+1}}\e {[i(ax+b)]}\}\)

We know,

\(\displaystyle (-1)^m=\e (-im\pi)=\e (im\pi)\)

\(\displaystyle (-i)^{m}=\e (-i\frac{m\pi}{2})\)

\(\displaystyle (i)^{m}=\e (i\frac{m\pi}{2})\),

then,

\(\displaystyle I_n=\frac{1}{2}\sum_{m=0}^{n}{\frac{\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\{\frac{(-1)^{m}\e {[-i(ax+b)]}}{(-i)^{m+1}}+\frac{(-1)^{m}\e {[i(ax+b)]}}{i^{m+1}}\}\)

\(\displaystyle =\frac{1}{2}\sum_{m=0}^{n}{\frac{\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\{\frac{\e (-im\pi)\e {[-i(ax+b)]}}{(-i)\e (-i\frac{m\pi}{2})}+\frac{\e (im\pi)\e {[i(ax+b)]}}{i\e (i\frac{m\pi}{2})}\}\)

\(\displaystyle =\sum_{m=0}^{n}{\frac{\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\{\frac{\e {[i(ax+b+\frac{m\pi}{2})]}-\e {[-i(ax+b+\frac{m\pi}{2})]}}{2i}\}\)

\(\displaystyle =\sum_{m=0}^{n}{\frac{\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\sin{(ax+b+\frac{m\pi}{2})}\).

Q.E.D.
 
Oct 2010
492
14
Changchun, China
Appendix I

\(\displaystyle E_n=\int x^n\e{(ax+b)}dx\)

\(\displaystyle =\frac{x^n}{a}\e(ax+b)+(-1)\frac{n}{a}\int x^{n-1}\e(ax+b)dx\)

\(\displaystyle =\frac{x^n}{a}\e(ax+b)+(-1)\frac{n}{a}E_{n-1}\)

\(\displaystyle =\frac{x^n}{a}\e(ax+b)+(-1)\times\frac{nx^{n-1}}{a^2}\e(ax+b)+(-1)^2\times\frac{n(n-1)}{a^2}E_{n-2}\)

.......

We know,

\(\displaystyle E_{n-m}=\int x^{n-m}\e{(ax+b)}dx=\frac{x^{n-m}}{a}\e(ax+b)+(-1)\frac{n}{a}E_{n-m-1}\)

Using the induction, finally,

\(\displaystyle E_n=\sum_{m=0}^{n}{\frac{(-1)^m\Gamma(n+1)x^{n-m}}{\Gamma(n-m+1)a^{m+1}}\e(ax+b)}\)
 

zaidalyafey

Math Team
Aug 2012
1,177
44
Sana'a , Yemen
stainburg said:
where \(\displaystyle \Gamma(.+1)=(.)!\)
Can you please explain what do you meant by that
 
Oct 2010
492
14
Changchun, China
zaidalyafey said:
Can you please explain what do you meant by that
Oops, sorry for my neglect.

\(\displaystyle \Gamma(z)\) is the Gamma function.

http://en.wikipedia.org/wiki/Gamma_function,

A very important property of the Gamma function is,

\(\displaystyle \Gamma(z+1)=z\Gamma(z)\)

If n is a positive integer,

\(\displaystyle \Gamma(n+1)=n!\), that's what I meant.
 

mathbalarka

Math Team
Mar 2012
3,871
86
India, West Bengal
zaidalyafey said:
Can you please explain what do you meant by that
Gamma function is the integral \(\displaystyle \int_{0}^{\infty} e^{-t} t^{z-1} dz\) which is a meromorphic extension of the factorial function.

Bohr Mollerups theorem states that this is a unique extension of the factorial function having a logarithmic convexity and hence, an analytic continuation.
 
Aug 2011
334
8
Nice work !
Nevertheless, the integral (x^n)cos(ax+b)dx can be expressed on a closed form, without series, thanks to the Incomplete Gamma function. Or thanks to the Generalised Fresnel functions (or Boehmer functions).
 
Oct 2010
492
14
Changchun, China
mathbalarka said:
Gamma function is the integral \(\displaystyle \int_{0}^{\infty} e^{-t} t^{z-1} dz\) which is a meromorphic extension of the factorial function.

Bohr Mollerups theorem states that this is a unique extension of the factorial function having a logarithmic convexity and hence, an analytic continuation.
Thanks, mathbalarka!
 
Oct 2010
492
14
Changchun, China
JJacquelin said:
Nice work !
Nevertheless, the integral (x^n)cos(ax+b)dx can be expressed on a closed form, without series, thanks to the Incomplete Gamma function. Or thanks to the Generalised Fresnel functions (or Boehmer functions).
Thanks J!

Lately, I teach myself the special functions finding it's rather beatiful when expanding a function with them!

I'm a beginner. I'll try the closed form you mentioned :p .Thanks again!
 
Oct 2010
492
14
Changchun, China
According to JJacquelin, we can change the series into a closed form by using the incomplete Gamma function:

\(\displaystyle \Gamma(s,x)=\int_x^{\infty} t^{s-1}\e{(-t)}dt\), which is the 'upper' incomplete Gamma function.

http://en.wikipedia.org/wiki/Incomplete_gamma_function

We now know the ordinary Gamma function is (Thanks to mathbalarka):

\(\displaystyle \Gamma(s)=\int_0^{\infty} t^{s-1}\e(-t)dt\).

Actually, there is relation between the ordinary one and 'upper' one( use the integration by parts and induction):

\(\displaystyle \Gamma(n+1,x)=\sum_{m=0}^{n}\frac{x^{m}}{\Gamma(m+1)}\Gamma(n+1)\e(-x)\), where n is a positive integer.

Back to the Appendix, we've got

\(\displaystyle E_n=\sum_{k=0}^{n}\frac{(-1)^k\Gamma(n+1)x^{n-k}}{\Gamma(n-k+1)a^{k+1}}\e (ax+b)\)(It means we count from 0 to n).

Now let \(\displaystyle m=n-k\) and substitute this into \(\displaystyle E_n\)(It means we count from n to 0):

\(\displaystyle E_n=\sum_{m=0}^{n}\frac{(-1)^{n-m}\Gamma(n+1)x^{m}}{\Gamma(m+1)a^{n-m+1}}\e (ax)\e(b)\)

\(\displaystyle =(-1)^na^{-(n+1)}\e(b)\sum_{m=0}^{n}\frac{\Gamma(n+1)(-ax)^{m}}{\Gamma(m+1)}\e [-(-ax)]\)

\(\displaystyle =-(-a)^{-(n+1)}\e(b)\Gamma(n+1,-ax)\).

Now we get the closed form of \(\displaystyle E_n\).

For \(\displaystyle I_n=\int x^n\cos(ax+b)dx\), we have

\(\displaystyle I_n=\frac{1}{2}\{\int x^n\e(-iax-ib)dx+\int x^n\e(iax+ib)dx\}\)

\(\displaystyle =-\frac{1}{2}\{(ia)^{-(n+1)}\e(-ib)\Gamma(n+1,iax)+(-ia)^{-(n+1)}\e(ib)\Gamma(n+1,-iax)\}\)

\(\displaystyle =\frac{i}{2a^{n+1}}\{\e[-i(b+\frac{n\pi}{2})]\Gamma(n+1,iax)-\e[i(b+\frac{n\pi}{2})]\Gamma(n+1,-iax)\}\)

That's one of the closed forms of \(\displaystyle I_n\). Thanks to JJacquelin.