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Definite Integral

bincybn

Member
Apr 29, 2012
36
Hii friends,


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




regards,
Bincy
 

themurgesh

New member
Mar 29, 2012
4
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:
\(\displaystyle \int_{0}^{1}\left(1-x\right)^{n}*e^{x}\, dx \;= \; [e\cdot n! - e \cdot \Gamma(n+1,1)] = e[\Gamma(n+1)-\Gamma(n+1,1)]\)
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
Hii friends,


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




regards,
Bincy
Let:

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

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

CB
 
Last edited:

CaptainBlack

Well-known member
Jan 26, 2012
890
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:
\(\displaystyle \int_{0}^{1}\left(1-x\right)^{n}*e^{x}\, dx \;= \; [e\cdot n! - e \cdot \Gamma(n+1,1)] = e[\Gamma(n+1)-\Gamma(n+1,1)]\)
Incomplete induction, not mathematical induction!

CB
 

bincybn

Member
Apr 29, 2012
36
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
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
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
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
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).

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
 
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bincybn

Member
Apr 29, 2012
36
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.
 
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CaptainBlack

Well-known member
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
 

bincybn

Member
Apr 29, 2012
36
Last edited:

CaptainBlack

Well-known member
Jan 26, 2012
890