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Hii friends,
\(\displaystyle \int_{0}^{1}\left(1-x\right)^{N}*e^{x}\, dx\)
regards,
Bincy
Let:Hii friends,
\(\displaystyle \int_{0}^{1}\left(1-x\right)^{N}*e^{x}\, dx\)
regards,
Bincy
Incomplete induction, not mathematical induction!this is what i have:
for n=0, \(\displaystyle \int_{0}^{1}\left(1-x\right)^{0}*e^{x}\, dx = e-1\)
for n=1, \(\displaystyle \int_{0}^{1}\left(1-x\right)^{1}*e^{x}\, dx = e-2\)
for n=2, \(\displaystyle \int_{0}^{1}\left(1-x\right)^{2}*e^{x}\, dx = 2e-5\)
for n=3, \(\displaystyle \int_{0}^{1}\left(1-x\right)^{3}*e^{x}\, dx = 6e-16\)
for n=4, \(\displaystyle \int_{0}^{1}\left(1-x\right)^{4}*e^{x}\, dx = 24e-65\)
and so on..
so you have two sequnces..
1,1,2,6,24... is simply n! for n=0,1,2....
1,2,5,16,65,... is \(\displaystyle e \cdot \Gamma(n+1,1)\) which is incomplete gamma function. For some reason I cannot paste the link here, please refer to the page mathworld.wolfram.com/BinomialSums.html
See equations 35 and 36
so you have:
\(\displaystyle \int_{0}^{1}\left(1-x\right)^{n}*e^{x}\, dx \;= \; [e\cdot n! - e \cdot \Gamma(n+1,1)] = e[\Gamma(n+1)-\Gamma(n+1,1)]\)
Integration by parts gives \(\displaystyle k*I_{k-1}-I_{k}=1 \)Let:
\( \displaystyle I_k=\int_0^1 (1-x)^k e^x \; dx\)
Then integration by parts gives: \(I_k+k I_{k-1}=1\)
CB
If themurgesh incomplete induction is correct, then if \(k\) is a non negative integer mathematical induction using the recurence of themurgesh's solution should work (though I am unhappy about the appearance of in incomplete gamma functions, in that I would rather avoid them if possible).Integration by parts gives \(\displaystyle k*I_{k-1}-I_{k}=1 \)
But how to solve this equation?
Is the ans \(\displaystyle \frac{k+1}{k-1} \) ?
thanks in advance.
Bincy
May I know what is wrong in my ans?
My ans. satisfies the recursive eqn.
I am also unhappy to include gamma function in my ans.
Oops. I didn't even bother abt the initial conditions.
Pls ignore this reply.
\[I_k=e \times n!+n! \sum_{k=0}^n \frac{1}{k!}\]
CB