Q. \[ \int \frac{\sin x}{x}\,dx \]

Q: What is \displaystyle \int_0^\infty \frac{\sin x}{x}\,dx ?

The Dirichlet integral gives \displaystyle \int_0^\infty \frac{\sin x}{x}\,dx = \frac{\pi}{2} .

Q: Is \displaystyle \int_0^a \frac{\sin x}{x}\,dx well-defined at x=0 ?

Yes, as an improper integral. Since \sin x \sim x near 0 , \frac{\sin x}{x}\to 1 . Thus \displaystyle \int_0^a \frac{\sin x}{x}\,dx = \operatorname{Si}(a) is finite.

Q: Series expansion for \frac{\sin x}{x} and integrating termwise?

\sin x = x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots . So \frac{\sin x}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots . Integrate: \operatorname{Si}(x)= x-\frac{x^3}{18}+\frac{x^5}{600}-\cdots .

Q: Differentiate \operatorname{Si}(x) .

By definition, \displaystyle \frac{d}{dx}\operatorname{Si}(x)=\frac{\sin x}{x} .

Q: What are key limiting values of \operatorname{Si}(x) ?

\operatorname{Si}(0)=0 . As x\to\infty , \operatorname{Si}(x)\to \frac{\pi}{2} . As x\to 0 , \operatorname{Si}(x)\sim x .

Answer

\( \displaystyle \int \frac{\sin x}{x}\,dx \) does not have an elementary antiderivative. It is expressed using the sine integral function \( \operatorname{Si}(x) \), defined by

\[
\operatorname{Si}(x)=\int_{0}^{x}\frac{\sin t}{t}\,dt.
\]

Thus, an antiderivative is

\[
\int \frac{\sin x}{x}\,dx=\operatorname{Si}(x)+C.
\]

Detailed Explanation

We want to compute the indefinite integral

\[ \int \frac{\sin x}{x}\, dx. \]

Step 1: Recognize what kind of integral this is.

The function \(\dfrac{\sin x}{x}\) does not have an elementary antiderivative (it cannot be expressed using a finite combination of polynomials, exponentials, logarithms, trig functions, and their inverses).

Step 2: Use the standard special function that matches this integrand.

A key related function is the cosine integral, defined by

\[
\mathrm{Ci}(x) = -\int_x^{\infty} \frac{\cos t}{t}\, dt,
\]

and one also uses the sine integral, defined by

\[
\mathrm{Si}(x) = \int_0^{x} \frac{\sin t}{t}\, dt.
\]

Step 3: Match the derivative of the sine integral.

By definition, \(\mathrm{Si}(x)\) satisfies

\[
\frac{d}{dx}\mathrm{Si}(x) = \frac{\sin x}{x}.
\]

Step 4: Conclude the antiderivative.

Since the derivative of \(\mathrm{Si}(x)\) is exactly \(\dfrac{\sin x}{x}\), we get

\[
\int \frac{\sin x}{x}\, dx = \mathrm{Si}(x) + C,
\]

where \(C\) is an arbitrary constant.

Final Answer:

\[
\int \frac{\sin x}{x}\, dx = \mathrm{Si}(x) + C.
\]

See full solution

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Calculus FAQ

Evaluate \( \displaystyle \int \frac{\sin x}{x}\,dx \).

No elementary antiderivative. It’s expressed as the sine integral: \( \displaystyle \int \frac{\sin x}{x}\,dx = \operatorname{Si}(x) + C \), where \( \operatorname{Si}(x)=\int_0^x \frac{\sin t}{t}\,dt \).

What is \( \displaystyle \int_0^\infty \frac{\sin x}{x}\,dx \)?

The Dirichlet integral gives \( \displaystyle \int_0^\infty \frac{\sin x}{x}\,dx = \frac{\pi}{2} \).

Is \( \displaystyle \int_0^a \frac{\sin x}{x}\,dx \) well-defined at \( x=0 \)?

Yes, as an improper integral. Since \( \sin x \sim x \) near \( 0 \), \( \frac{\sin x}{x}\to 1 \). Thus \( \displaystyle \int_0^a \frac{\sin x}{x}\,dx = \operatorname{Si}(a) \) is finite.

Series expansion for \( \frac{\sin x}{x} \) and integrating termwise?

\( \sin x = x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots \). So \( \frac{\sin x}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots \). Integrate: \( \operatorname{Si}(x)= x-\frac{x^3}{18}+\frac{x^5}{600}-\cdots \).

Differentiate \( \operatorname{Si}(x) \).

By definition, \( \displaystyle \frac{d}{dx}\operatorname{Si}(x)=\frac{\sin x}{x} \).

Compute \( \displaystyle \int_a^b \frac{\sin x}{x}\,dx \) in terms of special functions.

\( \displaystyle \int_a^b \frac{\sin x}{x}\,dx = \operatorname{Si}(b)-\operatorname{Si}(a) \).

What are key limiting values of \( \operatorname{Si}(x) \)?

\( \operatorname{Si}(0)=0 \). As \( x\to\infty \), \( \operatorname{Si}(x)\to \frac{\pi}{2} \). As \( x\to 0 \), \( \operatorname{Si}(x)\sim x \).

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