# Definite Integral

#### bincybn

##### Member
Hii friends,

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

regards,
Bincy

#### themurgesh

##### New member
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:

#### CaptainBlack

##### Well-known member
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
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
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 $$\displaystyle \frac{k+1}{k-1}$$ ?

Bincy

Last edited:

#### CaptainBlack

##### Well-known member
Integration by parts gives $$\displaystyle k*I_{k-1}-I_{k}=1$$

But how to solve this equation?

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

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

Last edited:

#### bincybn

##### Member
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.

Last edited:

#### CaptainBlack

##### Well-known member
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.

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

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

Last edited: