Part 1

(a) Evaluate $I_1$:

We are given: $I_n = \int x^n e^{-x^2} \, dx$ For $n = 1$, we have: $I_1 = \int x e^{-x^2} \, dx$

We can solve this by performing a u-substitution. Let: $u = -x^2 \quad \Rightarrow \quad du = -2x \, dx$

Thus, the integral becomes: $I_1 = \int x e^{-x^2} \, dx = -\frac{1}{2} \int e^u \, du = -\frac{1}{2} e^u + C = -\frac{1}{2} e^{-x^2} + C$

So, $I_1 = -\frac{1}{2} e^{-x^2} + C$

Part 2

(b) Use integration by parts to evaluate $I_3$:

For $n = 3$, we need to use integration by parts. Let: $I_3 = \int x^3 e^{-x^2} \, dx$

We choose:

• $u = x^2$, hence $du = 2x \, dx$,
• $dv = x e^{-x^2} \, dx$, hence $v = -\frac{1}{2} e^{-x^2}$.

Applying the integration by parts formula: $\int u \, dv = uv - \int v \, du$

This gives: $I_3 = -\frac{1}{2} x^2 e^{-x^2} + \frac{1}{2} \int 2x e^{-x^2} \, dx$

The second integral is $I_1$, so: $I_3 = -\frac{1}{2} x^2 e^{-x^2} + I_1$ Using the result from part (a) $I_1 = -\frac{1}{2} e^{-x^2} + C$, we get: $I_3 = -\frac{1}{2} x^2 e^{-x^2} - \frac{1}{2} e^{-x^2} + C$

Thus: $I_3 = -\frac{1}{2} (x^2 + 1) e^{-x^2} + C$

Part 3

(c) Use integration by parts and the result of part (b) to evaluate $I_5$:

For $n = 5$, we apply the same approach as in part (b): $I_5 = \int x^5 e^{-x^2} \, dx$

Using integration by parts: $I_5 = -\frac{1}{2} x^4 e^{-x^2} + 2 \int x^3 e^{-x^2} \, dx$

We already know from part (b) that $I_3 = -\frac{1}{2} (x^2 + 1) e^{-x^2} + C$, so substituting this: $I_5 = -\frac{1}{2} x^4 e^{-x^2} + 2 \left( -\frac{1}{2} (x^2 + 1) e^{-x^2} + C \right)$ Simplifying: $I_5 = -\frac{1}{2} x^4 e^{-x^2} - (x^2 + 1) e^{-x^2} + 2C$ Thus: $I_5 = -\frac{1}{2} (x^4 + 2x^2 + 2) e^{-x^2} + C$

Part 4

(d) Show that, in general, if $n$ is odd:

We are asked to show that: $I_n = -\frac{1}{2} e^{-x^2} p_{n-1}(x)$ where $p_{n-1}$ is an even polynomial of degree $n - 1$.

We already observed in parts (b) and (c) that when $n$ is odd, the result involves terms like $x^{n-1}, x^{n-3}, \dots$, which are even polynomials.

The correct general formula for odd $n$ is: $I_{2n+1} = -\frac{1}{2} e^{-x^2} (x^{2n} + \text{lower-degree terms})$

Thus, the correct answer is C: $I_{2n+1} = -\frac{1}{2} e^{-x^2} x^{2n} + n I_{2n-1}$

Part 5

(e) Argue that if $n$ is even, $I_n$ cannot be expressed in terms of elementary functions:

If $n$ is even, we encounter integrals that cannot be simplified using elementary functions.

The correct argument is: B. If $n$ is even, and a $u$-substitution is made for $x^2$, an integral of the form $\int e^{-u} \, du$ will always remain, which cannot be expressed in terms of elementary functions.

Questions:

1. Can you explain how the substitution method simplifies $I_1$?
2. Why does integration by parts work well for higher odd powers like $I_3$ and $I_5$?
3. How does the structure of $p_{n-1}(x)$ change for higher odd values of $n$?
4. Why can't $I_n$ be expressed in elementary functions when $n$ is even?
5. How does the recurrence relation in Part (d) help solve integrals for higher odd values of $n$?

Tip: Integration by parts is especially useful when combining polynomials and exponential functions like $e^{-x^2}$.