I have
$$j_n=\int_0^1 x^{2n} \sin(\pi x)dx.$$
How do I show that $$j_{n+1}= \frac{1}{\pi^2}(\pi- (2n+1)(2n+2)j_n)\, ?$$
I keep getting a recurring integration by parts and I can't simplify it. Please tell me where I'm going wrong.
Integration by parts gives
$$\int_0^1 x^k \sin(\pi x) dx = \frac{1}{\pi} + \frac{k}{\pi}\int_0^1 x^{k-1} \cos(\pi x) dx$$
and
$$\int_0^1 x^k \cos(\pi x) dx = - \frac{k}{\pi}\int_0^1 x^{k-1} \sin(\pi x) dx.$$
Applying these rules in succession to $j_{n+1}$ immediately gives
$$j_{n+1} = \frac{1}{\pi} - \frac{(2n+1)(2n+2)}{\pi^2} j_n$$
which clearly is equal to the desired formula.
Notice that this works for any real $n \gt -1$ (so that the integrals converge).
Brute-force attempts at integration produce error functions (Erf
), Dawson's integrals (DawsonF
), or generalized hypergeometric functions (HypergeometricPFQ
), and these do not readily simplify. Let's try something more elementary.
Having seen that the mathematical solution above requires two integrations by parts, we ought to be interested in the second derivative of $x^{2n+2}\sin(\pi x)$:
D[x^(2 n + 2) Sin[Pi x], {x, 2}]
$(2 n+1) (2 n+2) x^{2 n} \sin (\pi x)-\pi ^2 x^{2 n+2} \sin (\pi x) + \color{Red}{2 \pi (2 n+2) x^{2 n+1} \cos (\pi x)}$
The first two terms are exactly what we would be integrating when evaluating $(2n+1)(2n+2)j_{n} - \pi^2 j_{n+1}$ in order to check the equality we're trying to prove. The last term is a problem: how to make it go away? Well, still motivated by integrations by parts, we ought to examine a similar derivative,
D[x^(2 n + 1) Sin[Pi x], x]
$(2 n+1) x^{2 n} \sin (\pi x) + \color{Red}{\pi x^{2 n+1} \cos (\pi x)}$
The second term is a constant multiple of what we're trying to get rid of and the first one is not worrisome: it's proportional to the integrand for $j_n$. So, an appropriate linear combination of these two derivatives ought to clear things up a bit:
Expand[D[x^(2 n + 2) Sin[Pi x], {x, 2}] - 2 (2 + 2 n) D[x^(2 n + 1) Sin[Pi x], x] , x]
$-(2 n+1) (2 n+2) x^{2 n} \sin (\pi x)-\pi ^2 x^{2 n+2} \sin (\pi x)$
This can be stated in reverse: the indefinite integral of the result is what we differentiated, so let's just take one less derivative to find out what it is!
D[x^(2 n + 2) Sin[Pi x], x] - 2 (2 + 2 n) x^(2 n + 1) Sin[Pi x]
$\pi x^{2 n+2} \cos (\pi x)-(2 n+2) x^{2 n+1} \sin (\pi x)$
By the Fundamental Theorem of Calculus, we can find $-(2n+1)(2n+2)j_n - \pi^2 j_{n+1}$ by evaluating this antiderivative at the endpoints $0$ and $1$:
FullSimplify[% /. {x -> #} & /@ {0, 1} // Differences, Assumptions -> n > -1]
$\{-\pi \}$
That is, these Mathematica manipulations have demonstrated that
$$\eqalign{ -(2n+1)(2n+2)j_n - \pi^2 j_{n+1} &= \int_0^1 \left(-(2n+1)(2n+2)x^{2n}\sin{\pi x} - \pi^2 x^{2n+2}\sin(\pi x)\right) dx \\ &= \left(\pi x^{2 n+2} \cos (\pi x)-(2 n+2) x^{2 n+1} \sin (\pi x)\right)\mid_0^1 \\ &= -\pi. }$$
This obviously is algebraically equivalent to the desired result. But in the process we have obtained an indefinite integral for the integrands associated with this linear combination of $j_n$ and $j_{n+1}$.
I have more (mathematica) questions than answers.
Let us define
J[n_] := Integrate[ x^(2 n) Sin[Pi x], {x, 0, 1}]
(* (Pi*HypergeometricPFQ[{1 + n}, {3/2, 2 + n}, -Pi^2/4])/(2 + 2*n) *)
and
Clear[JJ];
JJ[n_] := JJ[n]=1/Pi^2 (Pi - (2 n ) (2 n -1) JJ[n - 1]) // Simplify
JJ[0] = J[0];
We have
Table[J[n] - JJ[n], {n, 0, 6}] // Simplify
(* {0,0,0,0,0,0,0} *)
which suggests the recursion is 'likely'
(Thanks to KennyColnago for pointing out an offset in the recursion)
My question would be: why
Table[J[n],{n,0,4}]//FindSequenceFunction
does not provide the recursion? And I don't understand why
Clear[j]; eqn = j[n] == b j[n - 1] + c /.
Solve[Table[J[n] == b J[n - 1] + c, {n, 1, 2}] // Simplify, {b,c}][[1]]
Table[eqn /. j -> J, {n, 1, 3}] // Simplify
produces
(* {True,True,False} *)
JJ[n_]
has JJ[n-1]
on the right-hand side, and so should have the $n$ there as $n-1$. That is, JJ[n_]:= 1/Pi^2 (Pi - (2 (n-1) + 2) (2 (n-1) + 1) JJ[n - 1])
. Thus I believe the recursion is correct.
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Oct 24, 2012 at 18:27
FindSequenceFunction
not long ago on this mathematica.stackexchange.com/questions/11042/… I was expecting it to work.
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Quite a manual process :
Solve the indefinite integral, then calculate the definite one :
indefInt[n_, x_] = Simplify[Integrate[x^(2 (n)) Sin[Pi x], x],
Assumptions -> {0 <= x <= 1, n \[Element] Integers}];
int[n_] = Assuming[n \[Element] Integers,
Limit[indefInt[n, x], x -> 1, Direction -> +1] -
Limit[indefInt[n, x], x -> 0, Direction -> -1]];
Now define an identity regarding the incomplete Gamma function :
recursion = Gamma[n_, z_] -> (n - 1) Gamma[n - 1, z] + z^(n - 1) Exp[-z];
Finally put everything together :
en = FullSimplify[int[n] /. recursion, Assumptions -> {n \[Element] Integers, n > 0}] ;
enp1 = FullSimplify[int[n + 1] /. recursion /. recursion /. recursion,
Assumptions -> {n \[Element] Integers, n > 0}];
FullSimplify[enp1 - 1/Pi^2 (Pi - (2 n + 1) (2 n + 2) en),
Assumptions -> {n \[Element] Integers, n > 0}]
(* 0 *)