Q. \[ \int e^{x^{2}}\,dx \]

Q: Find \int e^{-x^2}\,dx .

It is elementary in the error function. \int e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}\,\mathrm{erf}(x)+C.

Q: What substitution helps compute \int e^{x^2}\,dx ?

Let u=x^2, then du=2x\,dx. This doesn’t match dx, so it doesn’t simplify to elementary form; it leads to special functions (no elementary result).

Q: Why doesn’t \int e^{x^2}\,dx have an elementary form?

The integrand e^{x^2} grows too fast and the antiderivative is not representable using elementary functions. It is expressible via the imaginary error function \mathrm{erfi}(x).

Q: Differentiate \frac{\sqrt{\pi}}{2}\,\mathrm{erfi}(x) . Does it give e^{x^2}?

Yes. Since \frac{d}{dx}\mathrm{erfi}(x)=\frac{2}{\sqrt{\pi}}e^{x^2}, then by chain rule \frac{d}{dx}\left(\frac{\sqrt{\pi}}{2}\mathrm{erfi}(x)\right)=e^{x^2}.

Q: Give the power series for \int e^{x^2}\,dx .

Use e^{x^2}=\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}. Then \int e^{x^2}\,dx=C+\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)n!}.

Answer

We need \( \int e^{x^2}\,dx \). This integral has no elementary antiderivative (it cannot be expressed using basic functions). It can be written using the imaginary error function.

Result:

\[
\int e^{x^2}\,dx = \frac{\sqrt{\pi}}{2}\,\operatorname{erfi}(x)+C
\]
where \(C\) is a constant and \(\operatorname{erfi}(x)\) is the imaginary error function.

Detailed Explanation

We are asked to find the integral

\[
\int e^{x^2}\,dx.
\]

Step 1: Recognize whether there is an elementary antiderivative.

For many integrals of the form \( \int e^{x}\,dx \) or \( \int e^{ax+b}\,dx \), we can find an elementary antiderivative. However, here the exponent is \(x^2\), meaning we have \(e^{x^2}\), not \(e^{x}\) or \(e^{-x^2}\).

There is a key fact:

The functions \(e^{-x^2}\) and \(e^{x^2}\) have special behavior. The integral

\[
\int e^{-x^2}\,dx
\]

is expressed using the error function, which is a special function. The integral

\[
\int e^{x^2}\,dx
\]

also does not have an elementary antiderivative, and it is expressed using special functions as well.

Step 2: Use the imaginary error function.

The function called the imaginary error function, denoted \(\operatorname{erfi}(x)\), is defined so that it naturally appears when integrating \(e^{x^2}\).

Its definition is

\[
\operatorname{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{t^2}\,dt.
\]

Step 3: Relate the given integral to the definition.

From the definition, we can rewrite the integral of \(e^{x^2}\) as follows.

Start with the definition and isolate \(\int_0^x e^{t^2}\,dt\):

\[
\int_{0}^{x} e^{t^2}\,dt = \frac{\sqrt{\pi}}{2}\operatorname{erfi}(x).
\]

Step 4: Differentiate to confirm the antiderivative form.

The derivative of \(\operatorname{erfi}(x)\) is

\[
\frac{d}{dx}\operatorname{erfi}(x)=\frac{2}{\sqrt{\pi}}e^{x^2}.
\]

Now multiply both sides by \(\frac{\sqrt{\pi}}{2}\):

\[
\frac{\sqrt{\pi}}{2}\cdot \frac{d}{dx}\operatorname{erfi}(x)=\frac{\sqrt{\pi}}{2}\cdot \frac{2}{\sqrt{\pi}}e^{x^2}=e^{x^2}.
\]

This shows that

\[
\frac{d}{dx}\left(\frac{\sqrt{\pi}}{2}\operatorname{erfi}(x)\right)=e^{x^2}.
\]

Step 5: Write the general antiderivative.

Since the derivative of \(\frac{\sqrt{\pi}}{2}\operatorname{erfi}(x)\) is \(e^{x^2}\), an antiderivative is

\[
\int e^{x^2}\,dx=\frac{\sqrt{\pi}}{2}\operatorname{erfi}(x)+C,
\]

where \(C\) is an arbitrary constant.

Final Answer:

\[
\int e^{x^2}\,dx=\frac{\sqrt{\pi}}{2}\operatorname{erfi}(x)+C.
\]

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Calculus FAQ

Find \( \int e^{x^2}\,dx \).

There is no elementary antiderivative. Use the special function definition: \(\int e^{x^2}\,dx=\frac{\sqrt{\pi}}{2}\,\mathrm{erfi}(x)+C\), where \(\mathrm{erfi}(x)=-i\,\mathrm{erf}(ix)\).

Find \( \int e^{-x^2}\,dx \).

It is elementary in the error function. \(\int e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}\,\mathrm{erf}(x)+C\).

What substitution helps compute \( \int e^{x^2}\,dx \) ?

Let \(u=x^2\), then \(du=2x\,dx\). This doesn’t match \(dx\), so it doesn’t simplify to elementary form; it leads to special functions (no elementary result).

Why doesn’t \( \int e^{x^2}\,dx \) have an elementary form?

The integrand \(e^{x^2}\) grows too fast and the antiderivative is not representable using elementary functions. It is expressible via the imaginary error function \(\mathrm{erfi}(x)\).

Differentiate \( \frac{\sqrt{\pi}}{2}\,\mathrm{erfi}(x) \). Does it give \(e^{x^2}\)?

Yes. Since \(\frac{d}{dx}\mathrm{erfi}(x)=\frac{2}{\sqrt{\pi}}e^{x^2}\), then by chain rule \(\frac{d}{dx}\left(\frac{\sqrt{\pi}}{2}\mathrm{erfi}(x)\right)=e^{x^2}\).

Give the power series for \( \int e^{x^2}\,dx \).

Use \(e^{x^2}=\sum_{n=0}^{\infty}\frac{x^{2n}}{n!}\). Then \(\int e^{x^2}\,dx=C+\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)n!}\).

Solve ∫ e^x² dx step-by-step.
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