MHB Calculate Definite Integral with N Variable

[bincy]

Hii friends,[math]\int_{0}^{1}\left(1-x\right)^{N}*e^{x}\, dx[/math]

regards,
Bincy

 

[themurgesh]

Hii friends,

[math]\int_{0}^{1}\left(1-x\right)^{N}*e^{x}\, dx[/math]regards,
Bincy

this is what i have:for n=0, [math]\int_{0}^{1}\left(1-x\right)^{0}*e^{x}\, dx = e-1[/math]
for n=1, [math]\int_{0}^{1}\left(1-x\right)^{1}*e^{x}\, dx = e-2[/math]
for n=2, [math]\int_{0}^{1}\left(1-x\right)^{2}*e^{x}\, dx = 2e-5[/math]
for n=3, [math]\int_{0}^{1}\left(1-x\right)^{3}*e^{x}\, dx = 6e-16[/math]
for n=4, [math]\int_{0}^{1}\left(1-x\right)^{4}*e^{x}\, dx = 24e-65[/math]
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 [math]e \cdot \Gamma(n+1,1)[/math] 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 36so you have:
[math]\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)][/math]

 

[CaptainBlack]

Hii friends,[math]\int_{0}^{1}\left(1-x\right)^{N}*e^{x}\, dx[/math]

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

 

[CaptainBlack]

this is what i have:for n=0, [math]\int_{0}^{1}\left(1-x\right)^{0}*e^{x}\, dx = e-1[/math]
for n=1, [math]\int_{0}^{1}\left(1-x\right)^{1}*e^{x}\, dx = e-2[/math]
for n=2, [math]\int_{0}^{1}\left(1-x\right)^{2}*e^{x}\, dx = 2e-5[/math]
for n=3, [math]\int_{0}^{1}\left(1-x\right)^{3}*e^{x}\, dx = 6e-16[/math]
for n=4, [math]\int_{0}^{1}\left(1-x\right)^{4}*e^{x}\, dx = 24e-65[/math]
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 [math]e \cdot \Gamma(n+1,1)[/math] 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 36so you have:
[math]\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)][/math]

Incomplete induction, not mathematical induction!

CB

 

[bincy]

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 [math]k*I_{k-1}-I_{k}=1 [/math]

But how to solve this equation?

Is the ans [math] \frac{k+1}{k-1} [/math] ?thanks in advance.

Bincy

 

[CaptainBlack]

Integration by parts gives [math]k*I_{k-1}-I_{k}=1 [/math]

But how to solve this equation?

Is the ans [math] \frac{k+1}{k-1} [/math] ?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

 

[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.

 

[CaptainBlack]

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

 

[bincy]

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

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

 

[CaptainBlack]

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

Yes.

CB