The tutorials over at openAI are quite clear on how to use structured outputs. e.g. if we want a helper to perform an integral step by step, we can either use a “thinking model”, or use a regular model, but force the output be a list of steps:

from pydantic import BaseModel
from textwrap import dedent
from openai import OpenAI

client = OpenAI()
MODEL = "gpt-4o-mini"

integral_solver_prompt = """
    You are a calculus expert capable of solving indefinite integrals
    step-by-step. Given an integral problem, provide a detailed solution
    including each integration step and the final answer. For each step, provide
    the mathematical expression and an explanation of the rule or technique
    used.
"""


class IntegralSolution(BaseModel):
    class Step(BaseModel):
        explanation: str
        """A brief explanation of the step. Please use LaTeX for the
        mathematical expressions. use $..$ to enclose
        inline expressions and $$..$$ to enclose block expressions."""
        expression: str
        """
        A mathematical step in LaTeX.
        e.g., "$$ \int_0^1{2x + 3} dx = \int_0^1{2x} dx + \int_0^1{3} dx $$"
        """

    steps: list[Step]
    final_answer: str
    """
    e.g., "$$x^2 + 3x + C$$"
    """


def solve_integral(integral_problem: str) -> IntegralSolution:
    completion = client.beta.chat.completions.parse(
        model=MODEL,
        messages=[
            {"role": "system", "content": dedent(integral_solver_prompt)},
            {
                "role": "user",
                "content": f"Solve the following integral: {integral_problem}",
            },
        ],
        response_format=IntegralSolution,
    )
    return completion.choices[0].message.parsed

where we force the output to be a Pydantic model. Then, when asking

solve_integral("int[x^2 * sin(2x)dx] from 0 to 1")

We will use integration by parts to solve the integral. The formula for integration by parts is ( \int u , dv = uv - \int v , du ). We choose ( u = x^2 ) and ( dv = \sin(2x) , dx ).

u = x^2, , dv = \sin(2x) , dx.

Next, we compute ( du ) and ( v ). The derivative of ( u ) gives us ( du = 2x , dx ). To find ( v ), we integrate ( dv ): ( v = -\frac{1}{2} \cos(2x) ).

Now we apply the integration by parts formula: ( \int x^2 \sin(2x) , dx = uv - \int v , du ). Therefore, we have: ( -\frac{1}{2} x^2 \cos(2x) \Big|_0^1 + \frac{1}{2} \int 2x \cos(2x) , dx ).

Next, we first evaluate the boundary term: Evaluate ( -\frac{1}{2} x^2 \cos(2x) ) at the limits (0) to (1): When ( x = 1 ): -\frac{1}{2} \cdot 1^2 \cdot \cos(2) ), and when ( x = 0 ): -0. Thus we have: -\frac{1}{2} \cos(2).

Now we compute the remaining integral ( \int x \cos(2x) , dx ) using integration by parts again. Let ( u = x ) and ( dv = \cos(2x) , dx ). Then, we find ( du = dx ) and ( v = \frac{1}{2} \sin(2x) ).

u = x, , dv = \cos(2x) , dx. du = 2x , dx, , v = -\frac{1}{2} \cos(2x).

Now we apply the integration by parts formula: ( \int x^2 \sin(2x) , dx = uv - \int v , du ). Therefore, we have: ( -\frac{1}{2} x^2 \cos(2x) \Big|_0^1 + \frac{1}{2} \int 2x \cos(2x) , dx ).

Next, we first evaluate the boundary term: Evaluate ( -\frac{1}{2} x^2 \cos(2x) ) at the limits (0) to (1): When ( x = 1 ): -\frac{1}{2} \cdot 1^2 \cdot \cos(2) ), and when ( x = 0 ): -0. Thus we have: -\frac{1}{2} \cos(2).

Now we apply the integration by parts formula: ( \int x^2 \sin(2x) , dx = uv - \int v , du ). Therefore, we have: ( -\frac{1}{2} x^2 \cos(2x) \Big|_0^1 + \frac{1}{2} \int 2x \cos(2x) , dx ).

-\frac{1}{2} x^2 \cos(2x) \Big|_0^1 + \int x \cos(2x) , dx.

Next, we first evaluate the boundary term: Evaluate ( -\frac{1}{2} x^2 \cos(2x) ) at the limits (0) to (1): When ( x = 1 ): -\frac{1}{2} \cdot 1^2 \cdot \cos(2) ), and when ( x = 0 ): -0. Thus we have: -\frac{1}{2} \cos(2).

-\frac{1}{2} [1^2 \cos(2) - 0] = -\frac{1}{2} \cos(2).

Now we compute the remaining integral ( \int x \cos(2x) , dx ) using integration by parts again. Let ( u = x ) and ( dv = \cos(2x) , dx ). Then, we find ( du = dx ) and ( v = \frac{1}{2} \sin(2x) ).

u = x, , dv = \cos(2x) , dx.

Using integration by parts, we get: ( \int x \cos(2x) , dx = \frac{1}{2} x \sin(2x) \Big|_0^1 - \frac{1}{2} \int \sin(2x) , dx. ) Now we have to compute ( \int \sin(2x) , dx = -\frac{1}{2} \cos(2x) ).

= \frac{1}{2} x \sin(2x) \Big|_0^1 - \frac{1}{4} \cos(2x).

Evaluate ( \frac{1}{2} x \sin(2x) ) from 0 to 1: When ( x = 1 ): ( \frac{1}{2} \cdot 1 \cdot \sin(2) ), when ( x = 0 ): 0. Thus we have: ( \frac{1}{2} \sin(2) + \frac{1}{4} \cos(2) ).

= \frac{1}{2} \sin(2).

Now assemblng everything, we have: Total integral = ( -\frac{1}{2} \cos(2) + \left( \frac{1}{2} \sin(2) - \frac{1}{4} \cos(2) \right) ).

= -\frac{1}{2} \cos(2) + \frac{1}{2} \sin(2) - \frac{1}{4} \cos(2) = \frac{1}{2} \sin(2) - \frac{3}{4} \cos(2).

Finally, we find the definite integral from 0 to 1, which gives us the answer as a function of sine and cosine values at x=2.

Final result = \frac{1}{2} \sin(2) - \frac{3}{4} \cos(2).

\frac{1}{2} \sin(2) - \frac{3}{4} \cos(2)

Apart from the fact that the answer is wrong (and somehow gpt-4o-mini miss-spelled the word “assemblng” lol), it ignored the instruction to use LaTeX!

turns out, that under the hood openAI creates a json schema from the Pydantic model, but the docstrings of each field are not included. big ugh. Luckily, we can use the Field class together with descriptions to pass the relevant instructions to the model:

class IntegralSolution(BaseModel):
    class Step(BaseModel):
        explanation: str = Field(
            description="""
            A brief explanation of the step. Please use LaTeX for the
            mathematical expressions. use $..$ to enclose
            inline expressions and $$..$$ to enclose block expressions."""
        )
        expression: str = Field(
            description="""
            A mathematical step in LaTeX.
            e.g., "$$ \int_0^1{2x + 3} dx = \int_0^1{2x} dx + \int_0^1{3} dx $$"
        """
        )

    steps: list[Step]
    final_answer: str = Field(
        description="""
        e.g., "$$x^2 + 3x + C$$"
        """
    )

and when trying it out one more time:

output = solve_integral("int[x^2 * sin(2x)dx] from 0 to 1")

We need to solve the integral $$\int_0^1 x^2 \sin(2x) , dx$$. To do this, we will use integration by parts, which is suitable for products of functions like $x^2$ and $\sin(2x)$. The formula for integration by parts is given by $$\int u , dv = uv - \int v , du.$$

We choose our $u$ and $dv$ for integration by parts. Let $$u = x^2$$, then $$du = 2x , dx$$. For $$dv = \sin(2x) , dx$$, we integrate to find $$v$$. The integral of $\sin(2x)$ is $$v = -\frac{1}{2} \cos(2x)$$.

Now we apply the integration by parts formula that was stated earlier. We will calculate $$uv$$ at the bounds of the integral and substitute it back into our equation.

Evaluating the first part, we find the boundary values: when $x=1$, $$-\frac{1}{2} \cdot 1^2 \cdot \cos(2) = -\frac{1}{2} \cos(2)$$ and when $x=0$, it becomes $0$. Therefore, the evaluation gives us: $$-\frac{1}{2} \cos(2) - 0 = -\frac{1}{2} \cos(2).$$

Now we need to compute the remaining integral: $$\int_0^1 x \cos(2x) , dx$$. We apply integration by parts again. Let $$u = x$$ (hence $$du = dx$$) and $$dv = \cos(2x) , dx$$ (thus $$v = \frac{1}{2} \sin(2x)$$).

Evaluating the boundary terms $$\left[ \frac{1}{2} x \sin(2x) \right]_0^1$$, we have when $x=1$, it equals $$\frac{1}{2} \sin(2)$$ and when $x=0$, it equals $0$. Thus we have: $$\frac{1}{2} \sin(2)$$.

Now we need to compute the integral $$-\int_0^1 \frac{1}{2} \sin(2x) , dx$$. The integral of $\sin(2x)$ is $$-\frac{1}{2} \cos(2x)$$ hence: $$-\frac{1}{2} \left[-\frac{1}{2} \cos(2x) \right]_0^1 = -\frac{1}{4}[\cos(2) - 1].$$

Now combining all parts, from integration by parts, we have: $$I = -\frac{1}{2} \cos(2) + \frac{1}{2} \sin(2) + \frac{1}{4}[ ext{1} - \cos(2)].$$ This simplifies to: $$-\frac{1}{2} \cos(2) + \frac{1}{2} \sin(2) + \frac{1}{4} - \frac{1}{4} \cos(2)$$.

Thus we have our final answer for the integral $$\int_0^1 x^2 \sin(2x) , dx$$.

$$I = -\frac{3}{4} \cos(2) + \frac{1}{2} \sin(2) + \frac{1}{4}$$

much better! In the last step it messed up with the minus signs, $-\frac{1}{2} \left[-\frac{1}{2} \cos(2x) \right]_0^1$ where $-\times-=+$ not $-$, and apart from the oververbose explanation, it’s good :)