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- Thread starter bincybn
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- #1

- Mar 29, 2012

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Hii friends,

\(\displaystyle \int_{0}^{1}\left(1-x\right)^{N}*e^{x}\, dx\)

regards,

Bincy

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:

\(\displaystyle \int_{0}^{1}\left(1-x\right)^{n}*e^{x}\, dx \;= \; [e\cdot n! - e \cdot \Gamma(n+1,1)] = e[\Gamma(n+1)-\Gamma(n+1,1)]\)

Last edited:

- Jan 26, 2012

- 890

Let:Hii friends,

\(\displaystyle \int_{0}^{1}\left(1-x\right)^{N}*e^{x}\, dx\)

regards,

Bincy

\( \displaystyle I_k=\int_0^1 (1-x)^k e^x \; dx\)

Then integration by parts gives: \(k I_{k-1}-I_k=1\)

CB

Last edited:

- Jan 26, 2012

- 890

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:

\(\displaystyle \int_{0}^{1}\left(1-x\right)^{n}*e^{x}\, dx \;= \; [e\cdot n! - e \cdot \Gamma(n+1,1)] = e[\Gamma(n+1)-\Gamma(n+1,1)]\)

CB

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- #5

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

But how to solve this equation?

Is the ans \(\displaystyle \frac{k+1}{k-1} \) ?

thanks in advance.

Bincy

Last edited:

- Jan 26, 2012

- 890

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

The soliution without incomplete gamma functions, to which you can apply mathematical or complete induction is:

\[I_k=e \times n!-n! \sum_{k=0}^n \frac{1}{k!}\]

Your proposed answer cannot be right since is is wrong for all the cases where we know the answer.

CB

Last edited:

- Thread starter
- #7

- Jan 26, 2012

- 890

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.

See my previous post, it has been edited to include the actual solution which you can prove is the solution by induction.

CB

- Thread starter
- #9

\[I_k=e \times n!+n! \sum_{k=0}^n \frac{1}{k!}\]

CB

Thanks.. I got it. Instead of + it is -.

Last edited:

- Jan 26, 2012

- 890