Gemini vs DeepSeek for Math

The integral you are asking to solve is known as the Dirichlet Integral. It is a famous and deeply satisfying problem in calculus:

$$\int_{0}^{\infty} \frac{\sin(x)}{x} \, dx = \frac{\pi}{2}$$

Because the antiderivative of $\frac{\sin(x)}{x}$ (the sinc function) cannot be expressed in terms of elementary functions, we cannot solve this using the standard Fundamental Theorem of Calculus. Instead, we can use a beautiful technique popularized by physicist Richard Feynman, known as Differentiation Under the Integral Sign (or Feynman's Trick).

Here is the complete step-by-step reasoning.

Step 1: Introduce a Parameter

We define a new function $I(t)$ by introducing a damping factor, $e^{-tx}$, into the integral, where $t \ge 0$:

$$I(t) = \int_{0}^{\infty} \frac{\sin(x)}{x} e^{-tx} \, dx$$

Notice that our original integral is simply $I(0)$. We have transformed the problem into finding the value of this function when the parameter $t = 0$.

Step 2: Differentiate Under the Integral Sign

We take the derivative of $I(t)$ with respect to $t$. Thanks to Leibniz's Rule, we can move the derivative inside the integral sign. Since the integral is with respect to $x$, we treat $x$ as a constant while differentiating with respect to $t$:

$$I'(t) = \frac{d}{dt} \int_{0}^{\infty} \frac{\sin(x)}{x} e^{-tx} \, dx$$ $$I'(t) = \int_{0}^{\infty} \frac{\partial}{\partial t} \left( \frac{\sin(x)}{x} e^{-tx} \right) \, dx$$

The derivative of $e^{-tx}$ with respect to $t$ is $-x e^{-tx}$. Notice how the $-x$ beautifully cancels out the problematic $x$ in the denominator:

$$I'(t) = \int_{0}^{\infty} \frac{\sin(x)}{x} (-x e^{-tx}) \, dx$$ $$I'(t) = -\int_{0}^{\infty} \sin(x) e^{-tx} \, dx$$

Step 3: Evaluate the New Integral

We now have a standard integral that can be solved using Integration by Parts (twice) or by using Euler's formula. Solving $\int e^{-tx} \sin(x) , dx$ yields:

$$\int e^{-tx} \sin(x) \, dx = \frac{-e^{-tx}(t \sin(x) + \cos(x))}{t^2 + 1}$$

Evaluating this from $0$ to $\infty$:

1. As $x \to \infty$, the term $e^{-tx}$ goes to $0$ (since $t > 0$), making the whole expression $0$.
2. At $x = 0$, $\sin(0) = 0$ and $\cos(0) = 1$, so the expression evaluates to $\frac{-1(0 + 1)}{t^2 + 1} = -\frac{1}{t^2 + 1}$.

Subtracting the lower bound from the upper bound:

$$I'(t) = - \left( 0 - \left( -\frac{1}{t^2 + 1} \right) \right) = -\frac{1}{t^2 + 1}$$

Step 4: Integrate Back to Find $I(t)$

Now that we have $I'(t)$, we integrate it with respect to $t$ to recover $I(t)$:

$$I(t) = \int -\frac{1}{t^2 + 1} \, dt$$

This is a standard arctangent integral:

$$I(t) = -\arctan(t) + C$$

Step 5: Solve for the Constant $C$

To find the constant of integration $C$, we look at the behavior of the original parametric function as $t \to \infty$:

$$\lim_{t \to \infty} I(t) = \lim_{t \to \infty} \int_{0}^{\infty} \frac{\sin(x)}{x} e^{-tx} \, dx$$

As $t$ becomes infinitely large, $e^{-tx}$ approaches $0$ everywhere except at $x=0$, forcing the whole integral to converge to $0$. Therefore:

$$\lim_{t \to \infty} I(t) = 0$$

Substitute $t \to \infty$ into our arctangent equation:

$$0 = -\lim_{t \to \infty} \arctan(t) + C$$

Since we know that $\arctan(\infty) = \frac{\pi}{2}$:

$$0 = -\frac{\pi}{2} + C \implies C = \frac{\pi}{2}$$

Now we have the full formula for our parametric function:

$$I(t) = \frac{\pi}{2} - \arctan(t)$$

Step 6: Find the Original Integral

Recall from Step 1 that our original integral is exactly $I(0)$. We simply plug $t = 0$ into our formula:

$$I(0) = \frac{\pi}{2} - \arctan(0)$$

Since $\arctan(0) = 0$, we arrive at our final answer:

$$\int_{0}^{\infty} \frac{\sin(x)}{x} \, dx = \frac{\pi}{2}$$

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Alternative Methods

While Feynman's Trick is highly intuitive, the Dirichlet Integral can be solved through other advanced mathematical avenues.

1. The Laplace Transform Method

The function $\frac{\sin(x)}{x}$ is practically begging for a Laplace transform. A known property of the Laplace Transform states:

$$\int_{0}^{\infty} f(x) g(x) \, dx = \int_{0}^{\infty} (\mathcal{L}f)(s) (\mathcal{L}^{-1}g)(s) \, ds$$

If we let $f(x) = \sin(x)$ and $g(x) = \frac{1}{x}$, we get:

• $\mathcal{L}{\sin(x)}(s) = \frac{1}{s^2 + 1}$
• $\mathcal{L}^{-1}{\frac{1}{x}}(s) = 1$

Applying this to our integral:

$$\int_{0}^{\infty} \frac{\sin(x)}{x} \, dx = \int_{0}^{\infty} \frac{1}{s^2 + 1} \cdot 1 \, ds$$

This directly gives us the arctangent integral from $0$ to $\infty$:

$$\left[ \arctan(s) \right]_{0}^{\infty} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$

2. Contour Integration (Complex Analysis)

Using Cauchy's Residue Theorem, we can evaluate this by extending the integral to the complex plane:

$$\int_{-\infty}^{\infty} \frac{\sin(x)}{x} \, dx = \text{Im} \left( \oint \frac{e^{iz}}{z} \, dz \right)$$

By drawing a contour along the real axis (with an infinitesimally small semi-circular detour over the singularity at $z=0$) and closing the loop with a giant semicircle in the upper half-plane, applying Jordan's Lemma tells us the outer arc vanishes to zero. Evaluating the residue precisely at the "dent" around $z=0$ ultimately yields $\pi$ for the integral from $-\infty$ to $\infty$. Since $\frac{\sin(x)}{x}$ is an even function, the integral from $0$ to $\infty$ is merely half of it, yielding $\frac{\pi}{2}$.

Alright, let’s solve step-by-step:

We want to evaluate
$$I = \int_{0}^{\infty} \frac{\sin x}{x} \, dx$$

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1. Introducing a parameter for differentiation under the integral sign

Consider
$$F(t) = \int_{0}^{\infty} \frac{\sin x}{x} \, e^{-tx} \, dx, \quad t \ge 0$$
We note that $F(0)$ is our desired integral, and $\lim_{t \to \infty} F(t) = 0$.

Differentiate $F(t)$ w.r.t. $t$:

$$F'(t) = \int_{0}^{\infty} \frac{\sin x}{x} \cdot (-x) e^{-tx} \, dx$$
$$F'(t) = -\int_{0}^{\infty} \sin x \cdot e^{-tx} \, dx$$

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2. Evaluate $\int_{0}^{\infty} e^{-tx} \sin x \, dx$

This is a standard Laplace transform of $\sin x$:

Recall formula
$$\mathcal{L}\{\sin(ax)\}(s) = \frac{a}{s^2 + a^2}$$

Here $a = 1$, $s = t$:

$$\int_{0}^{\infty} e^{-tx} \sin x \, dx = \frac{1}{t^2 + 1}$$

So
$$F'(t) = -\frac{1}{t^2 + 1}$$

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3. Integrate $F'(t)$ to get $F(t)$

$$F(t) = -\int \frac{1}{t^2 + 1} \, dt = -\arctan(t) + C$$

We know $\lim_{t \to \infty} F(t) = 0$:

$$0 = -\lim_{t \to \infty} \arctan(t) + C \quad \Rightarrow \quad 0 = -\frac{\pi}{2} + C$$
$$C = \frac{\pi}{2}$$

So
$$F(t) = \frac{\pi}{2} - \arctan(t) = \arctan\left(\frac{1}{t}\right), \quad t > 0$$

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4. Find $F(0)$

We have $F(t) = \int_{0}^{\infty} \frac{\sin x}{x} e^{-tx} \, dx$.

We can't directly plug $t=0$ into $\arctan(1/t)$, but we can take limit $t \to 0^+$:

$$\lim_{t \to 0^+} F(t) = \lim_{t \to 0^+} \arctan\left(\frac{1}{t}\right) = \frac{\pi}{2}$$

But we must check that
$$F(0) = \int_{0}^{\infty} \frac{\sin x}{x} \, dx$$
is equal to $\lim_{t \to 0^+} F(t)$.

We have $e^{-tx} \le 1$ for $t\ge 0$, and $\int_{0}^{\infty} \frac{\sin x}{x} \, dx$ converges (by Dirichlet's test or alternating series after breaking into $\pi$-intervals). Then by dominated convergence theorem, $F$ is continuous at $t=0$. Hence
$$F(0) = \lim_{t \to 0^+} F(t) = \frac{\pi}{2}$$

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5. Conclusion

$$\int_{0}^{\infty} \frac{\sin x}{x} \, dx = \frac{\pi}{2}$$

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Final answer: $$\boxed{\frac{\pi}{2}}$$

Techniques used:

1. Introduction of a convergence factor $e^{-tx}$
2. Differentiation under the integral sign (Feynman’s trick)
3. Laplace transform of sine
4. Dominated convergence theorem to justify $F$ continuous at $t=0$