Integration by parts MCQ

Question

Let $I_n=\int_{0}^{\infty} x^n e^{-x}~dx$, where $n$ is some positive integer. Then $I_n$ equals:

$\text{A) } n!-nI_{n-1}$

$\text{B) } n!+nI_{n-1}$

$\text{C) } nI_{n-1}$

$\text{D) } \text{none of these}$

Please help with this, the correct answer is $\text{C}$. Is there a fast method to solve such problems?

Answer 1

Substitute for $u = x^{n+1}$. Then we get, $$ I = \int_{0}^{\infty} x^n e^{-x} dx =\int_{0}^{\infty} \frac{1}{n+1} e^{-u^{\frac{1}{n+1}}} du$$ This is the incomplete gamma function. Thus, finally $$ I = -(\Gamma(n+1, \infty)-\Gamma(n+1,0)) = \Gamma(n+1,0) = \Gamma(n+1)=(n+1)!$$ This is satisfied by the choice $(c)$. Hope it helps.

Answer 2

Consider this function:

$$f_n(y)=\int_0^\infty x^ne^{-yx}\ dx=\frac{f_n(1)}{y^{n+1}}$$

Use u-substitution to show this. If we differentiate w.r.t. $y$ one time, we get

$$f'_n(y)=-\int_0^\infty x^{n+1}e^{-yx}\ dx=-f_{n+1}(y)\tag{LHS}$$

$$\frac d{dy}\frac{f_n(1)}{y^{n+1}}=\frac{-(n+1)f_n(1)}{y^{n+2}}\tag{RHS}$$

Set these equal to get

$$f_{n+1}(y)=\frac{(n+1)f_n(1)}{y^{n+2}}$$

And at $y=1$, we get

$$f_{n+1}(1)=(n+1)f_n(1)$$

where

$$f_n(1)=\int_0^\infty x^ne^{-x}\ dx$$

Or rewriting this, we have

C)$$I_n=nI_{n-1}$$

Answer 3

Let $$u = x^n, \quad du = nx^{n-1} \, dx, \\ dv = e^{-x} \, dx, \quad v = -e^{-x}.$$ This immediately gives $$I_n = \lim_{t \to \infty} \left[-x^n e^{-x}\right]_{x=0}^t + n \int_{x=0}^\infty x^{n-1} e^{-x} \, dx = \lim_{x \to \infty} -x^n e^{-x} + n I_{n-1}.$$ For any finite positive $n$, the limit approaches $0$ and the result is proven.

Answer 4

$\displaystyle I_{n} = \int^{\infty}_{0}x^n \cdot e^{-x}dx = -x^n\cdot e^{-x}|_{0}^{\infty}+n\int^{\infty}_{0}x^{n-1}\cdot e^{-x}dx = nI_{n-1}$

So using Recursively $$\displaystyle I_{n} = (n-0)(n-1)I_{n-2}=n(n-1)(n-2)I_{n-3}=n(n-1)(n-2) \cdot \cdot \cdot \cdot (n-(n-1))I_{n-(n)}$$

So $$I_{n}=n!\cdot I_{0} = -n!(e^{-x})|^{\infty}_{0} = n!$$