Question

Solve the following integral-
\[\int\limits_0^\infty {\mathop x\nolimits^n } \mathop e\nolimits^{ - x} dx\] (n is positive integer) is equal to
A) n!
B) (n-1)!
C) (n-2)!
D) (n+1)!

Answer 1

Hint: The integral of type \[\int\limits_0^\infty {\mathop x\nolimits^{n - 1} } \mathop e\nolimits^{ - x} dx\] is solved by using gamma function i.e.
\[\int\limits_0^\infty {\mathop x\nolimits^{n - 1} } \mathop e\nolimits^{ - x} dx = \left| \!{\overline {\,
 n \,}} \right. \]
Here, are some standard result for gamma function $\left| \!{\overline {\,
 n \,}} \right. $:
$
  \left| \!{\overline {\,
 {n + 1} \,}} \right. = n! \\
  \left| \!{\overline {\,
 {\dfrac{1}{2}} \,}} \right. = \sqrt \pi \\
  \left| \!{\overline {\,
 1 \,}} \right. = 1 \\
 $

Complete step-by-step answer:
Step 1: Given integral \[I = \int\limits_0^\infty {\mathop x\nolimits^n } \mathop e\nolimits^{ - x} dx\].
In order to use gamma function: \[\int\limits_0^\infty {\mathop x\nolimits^{n - 1} } \mathop e\nolimits^{ - x} dx = \left| \!{\overline {\,
 n \,}} \right. \]

Step 2: Reduce the given integral accordingly to use gamma function.
Here, the power of the variable, x is in the form $(n - 1)$.
Therefore, reduce the power of variable, x in given integral in the form $(n - 1)$.
The power of variable, x in given integral = n
Adding +1 and -1 to the power won’t change the function.
$ \Rightarrow n - 1 + 1$
$ \Rightarrow \left[ {(n + 1) - 1} \right]$
Thus, the power of variable, x in given integral can be written as $\left[ {\left( {n + 1} \right) - 1} \right]$

Step 3: Solve integral using gamma function
Hence, \[I = \int\limits_0^\infty {\mathop x\nolimits^{\left[ {\left( {n + 1} \right) - 1} \right]} } \mathop e\nolimits^{ - x} dx\].
By gamma function
\[I = \int\limits_0^\infty {\mathop x\nolimits^{\left[ {\left( {n + 1} \right) - 1} \right]} } \mathop e\nolimits^{ - x} dx = \left| \!{\overline {\,
 {n + 1} \,}} \right. \]
$ \Rightarrow \left| \!{\overline {\,
 {n + 1} \,}} \right. = n!$
Final answer: The integral \[\int\limits_0^\infty {\mathop x\nolimits^n } \mathop e\nolimits^{ - x} dx = n!\].
Thus the correct option is (A).

Note:To use gamma function, the limits of the integral must be from $0 \to \infty $.
The question of type \[\int\limits_0^\infty {\mathop e\nolimits^{ - \mathop x\nolimits^2 } } dx\]can also be solved by the help of gamma function.
\[I = \int\limits_0^\infty {\mathop e\nolimits^{ - \mathop x\nolimits^2 } } dx\]
Take \[\mathop x\nolimits^2 = t\]
$\therefore x = \sqrt t $
On differentiating both sides
$
  2xdx = dt \\
   \Rightarrow dx = \dfrac{{dt}}{{2x}} \\
 $
$
   \Rightarrow dx = \dfrac{{dt}}{{2\sqrt t }} \\
   \Rightarrow dx = \dfrac{1}{2}\mathop t\nolimits^{ - \dfrac{1}{2}} {\text{ }}dt \\
 $
Change of limits:
When, $x \to 0 \Rightarrow t \to 0$
           $x \to \infty \Rightarrow t \to \infty $
On substitution integral becomes
$I = \dfrac{1}{2}\int\limits_0^\infty {\mathop t\nolimits^{ - \dfrac{1}{2}} } \mathop e\nolimits^{ - t} {\text{ }}dt$
In order to use gamma function: \[\int\limits_0^\infty {\mathop x\nolimits^{n - 1} } \mathop e\nolimits^{ - x} dx = \left| \!{\overline {\,
 n \,}} \right. \]
Here, the power of the variable, x is in the form $(n - 1)$.
Therefore, reduce the power of the variable, t in above integral in the form $(n - 1)$.
The power of variable, t in above integral = $ - \dfrac{1}{2}$
Adding +1 and -1 to the power won’t change the function.
$ \Rightarrow - \dfrac{1}{2} - 1 + 1$
$ \Rightarrow \left( {\dfrac{1}{2} - 1} \right)$
Thus, the power of variable, t in above integral can be written as $\left( {\dfrac{1}{2} - 1} \right)$
Hence, $I = \dfrac{1}{2}\int\limits_0^\infty {\mathop t\nolimits^{\left( {\dfrac{1}{2} - 1} \right)} } \mathop e\nolimits^{ - t} {\text{ }}dt$
By gamma function
$ \Rightarrow \dfrac{1}{2}\left| \!{\overline {\,
 {\dfrac{1}{2}} \,}} \right. $
We know, $\left| \!{\overline {\,
 {\dfrac{1}{2}} \,}} \right. = \sqrt \pi $
$\because I = \dfrac{{\sqrt \pi }}{2}$