Consider the improper integral I(a,b)=∫_0^(∞) x^a e^(-b x) dx, where a>-1 and b>0.

Consider the improper integral $I (a, b) = \int_{0}^{\infty} x^{a} e^{- b x} d x$ , where $a > - 1$ and $b > 0$ .

hard 5 marks Class 12 · Mathematics · Integrals

Answer

0.085

The correct answer is $0.085$ — Rounded to three decimal places, $I = 0.085$ .

For $x \to 0^{+}$ , $x^{a}$ behaves like $x^{a}$ ; the integrand is integrable near $0$ iff $a > - 1$ .
For $x \to \infty$ , the factor $e^{- b x}$ dominates any power of $x$ , so the integral always converges for $b > 0$ .
Hence the integral converges exactly when $a > - 1$ (with $b > 0$ already given).
Recognise the definition of the Gamma function: $Γ (s) = \int_{0}^{\infty} t^{s - 1} e^{- t} d t$ for $s > 0$ .
Let $t = b x \Rightarrow x = \frac{t}{b}, d x = \frac{d t}{b}$ .
Substitute into $I (a, b)$ : $I (a, b) = \int_{0}^{\infty} (\frac{t}{b})^{a} e^{- t} \frac{d t}{b} = \frac{1}{b ^{a + 1}} \int_{0}^{\infty} t^{a} e^{- t} d t$ .
The remaining integral is $Γ (a + 1)$ . Thus $I (a, b) = \frac{Γ ( a + 1 )}{b ^{a + 1}}$ .
Now set $a = \frac{3}{2}$ , $b = 3$ . Compute $Γ (\frac{5}{2})$ . Using $Γ (\frac{5}{2}) = \frac{3}{2} \cdot \frac{1}{2} π = \frac{3}{4} π$ .
Numerically, $π \approx 1.77245385$ , so $Γ (\frac{5}{2}) \approx \frac{3}{4} \times 1.77245385 = 1.3293404$ .
Compute $b^{a + 1} = 3^{\frac{5}{2}} = 3^{2} 3 = 9 \times 1.7320508 = 15.588457$ .
Finally, $I = \frac{1.3293404}{15.588457} \approx 0.0853$ .
Rounded to three decimal places, $I = 0.085$ .

Common mistake: Forgetting the condition

a > - 1

or mixing up

a + 1

with

a

in the Gamma function argument.

Concept tested: Gamma function and convergence of improper integrals