Integral with parameter

[Camillo]

Calculate for $ AA lambda in RR $ , the following integral :


$G(lambda) = int_(-1)^1 (e^(lambdax)/(2-x))*dx$

[luca.barletta]

First, consider the following:

$e^(lambdax)=sum_(n=0)^(+infty) (lambdax)^n/(n!)$

Substituting into the integral:

$int_(-1)^1 e^(lambdax)/(2-x) dx=int_(-1)^1 1/(2-x) sum_(n=0)^(+infty) (lambdax)^n/(n!) dx=
$=sum_(n=0)^(+infty) (lambda)^n/(n!) int_(-1)^1 (x^n)/(2-x) dx$

Now, compute one "per parti" step of the previous integral:

$int_(-1)^1 (x^n)/(2-x) dx=[-x^n*ln(2-x)]_(-1)^1+int_(-1)^1 nx^(n-1)ln(2-x)dx=$
$=(-1)^nln3+int_(-1)^1 nx^(n-1)ln(2)dx+int_(-1)^1 nx^(n-1)ln(1-x/2)dx=$
$=(-1)^nln3+ln(2)(1-(-1)^n)+int_(-1)^1 nx^(n-1)ln(1-x/2)dx$

For the last integral I consider the equality:

$ln(1-x/2)=sum_(k=1)^(+infty) (-1)^(k+1)/k (-x/2)^k$

and substituting it in the integral:

$int_(-1)^1 nx^(n-1)ln(1-x/2)dx=-sum_(k=1)^(+infty) 1/(k2^k)int_(-1)^1 nx^(n-1)x^kdx=$
$=-sum_(k=1)^(+infty) n/(k2^k(n+k))(1-(-1)^(n+k))$

At this point we have:

$G(lambda)=sum_(n=0)^(+infty) (lambda)^n/(n!)*((-1)^nln3+ln(2)(1-(-1)^n)-sum_(k=1)^(+infty) n/(k2^k(n+k))(1-(-1)^(n+k)))$

Finally we can consider the following 4 cases:

$G(lambda)=sum_(n=0)^(+infty) (lambda)^n/(n!)*{((ln3),(n even, k even)),((ln3-sum_(k) n/(k2^(k-1)(n+k))),(n even, k odd)),((-ln3+2ln2-sum_(k) n/(k2^(k-1)(n+k))),(n odd, k even)),((-ln3+2ln2),(n odd,k odd)):}$

[nicola de rosa]

First, consider the following:
$1/(2-x)=1/2*1/(1-x/2)=1/2*sum_{n=0}^{infty)(x/2)^n$. So
$G(lambda)=sum_{n=0}^{infty)2^(-(n+1))*int_{-1}^{1}x^n*e^(lambda*x)dx$
Now we consider that $e^(lambdax)=sum_(k=0)^(+infty) (lambda*x)^k/(k!)$, therefore

$G(lambda)=sum_{n=0}^{infty)sum_{k=0}^{infty)2^(-(n+1))*(lambda)^k/(k!)*int_{-1}^{+1}x^(n+k)dx$
Now $int_{-1}^{+1}x^(n+k)dx=(1-(-1)^(n+k+1))/(n+k+1)$
So
$G(lambda)=sum_{n=0}^{infty)sum_{k=0}^{infty)2^(-(n+1))*(lambda)^k/(k!)*(1-(-1)^(n+k+1))/(n+k+1)$

[Camillo]

I will post shortly the solution I have ; then it will be interesting to compare if the various solutions, which looks like quite different , are the same ...

[Camillo]

If $lambda = 0 $ it is immediate to obtain : $G(0) = ln3 $.
For $ lambda ne 0 $, the function to be integrated : $e^(lambda.x)*(2-x)^(-1) $ is elementary, but it has no elementary
primitive.

Now for $-2<x<2 $ we get :
$1/(2-x) =(1/2)*1/(1-(x/2)) = (1/2)sum_(n=0)^(+oo)(x/2)^n $ .
The series is uniformly convergent in the integration interval $(-1<=x<=1)$.
Since $e^(lambdax)$ is,in the same interval, a limited function $(0 <e^(lambdax)<=e^|lambda| )$ , we can integrate by series the original function and we get :
$ G(lambda) = int_(-1)^(1) sum_(n=0)^(+oo)e^(lambdax)*x^n*dx/2^(n+1)=sum_(n=0)^(+oo)1/2^nint_(-1)^(1)e^(lambdax)*x^n*dx/2$.
We have therefore to calculate the integrals : $ I_n ( lambda) = 1/2int_(-1)^(1) e^(lambdax)*x^n*dx $, $AA n in N$.
We can integrate by parts ,$AA n$.Or more rapidly we have (remember derivation under the sign of integral) that :

$d/(dlambda) int_(-1)^(1)e^(lambdax)*x^n*dx =int_(-1)^(1)(del/(dellambda))e^(lambdax)*x^n*dx=int_(-1)^(1)x*e^(lambdax)*x^n*dx=int_(-1)^(1)e^(lambdax)*x^(n+1)*dx$.
Therefore :
$I_(n+1)(lambda) = I_n '(lambda) $, $n= 0,1,2 ,..$.
Consequently :
$I_1(lambda) = I_0'(lambda)$
$I_2(lambda) =I_1 '(lambda)=I_0''(lambda) ,....,I_n(lambda)=I_0^(n)(lambda)$.
We have also : $ I_0(lambda)=1/2int_(-1)^(1)e^(lambdax)*dx= (Sh lambda)/lambda $
and as a consequence :

$I_n(lambda) =d^n/(dlambda^n)(Sh lambda)/lambda$
Finally the result which appears quite concise :

$G(lambda) = sum_(n=0)^(+oo)1/2^n(d^n/(dlambda^n)(Shlambda)/lambda) $