# Derivative of $f(x)=\int^{x^2}_0 \frac{\sin(t)}{t}dt$

Let $f(x)=\int^{x^2}_0 \frac{\sin(t)}{t}dt$. Find $f'(x)$.

Is that integral undefined/nonexistant, or just impossible to integrate?

In this case does $f'(x)$ exist and can be solved normally?

Is it correct that $f'(x)=\frac{\sin(x^2)}{x^2}2x$?

• Fundamental theorem of calculus. – Jacky Chong Oct 9 '16 at 1:47
• $\int_0^{x^2} \frac{\sin(t)}{t}\,dt$ is integrable for all $x$. – Mark Viola Oct 9 '16 at 1:49
• for all x? Does that mean that t is never 0 for any x? – Arne Morten Oct 9 '16 at 20:49

Notice that if we call $$g(u) = \int^{u}_0 \frac{sin(t)}{t}dt$$ then we know that $g'(u)=\frac{\sin u}{u}$

But $f(x)=g(x^2)$. Then you apply the chain rule to get your result.

Is that integral undefined/nonexistant

The integral is perfectly defined. It just cannot be expressed in closed form, in terms of elementary functions. But that does not matter at all.

By the fundamental theorem of calculus, if $F$ is an antiderivative of $f$, then $$\int_{\ell(x)}^{u(x)}f(t)\text{ d}t = F(u(x)) - F(\ell(x))\text{.}$$ It follows that, after applying the chain rule, $$\dfrac{\text{d}}{\text{d}x}\int_{\ell(x)}^{u(x)}f(t)\text{ d}t = F^{\prime}(u(x))u^{\prime}(x) - F^{\prime}(\ell(x))\ell^{\prime}(x)\text{.}$$ but since $F$ is an antiderivative of $f$, it follows that $F^{\prime} = f$, hence $$\dfrac{\text{d}}{\text{d}x}\int_{\ell(x)}^{u(x)}f(t)\text{ d}t = f(u(x))u^{\prime}(x) - f(\ell(x))\ell^{\prime}(x)\text{.}$$

This rule obviously requires many conditions (notice, in this case, $f(\ell(x))$ is undefined), so we use a slight modification here: observe that if $F$ is an antiderivative of $\dfrac{\sin(x)}{x}$, $F(0)$ is obviously a constant; then, $$\dfrac{\text{d}}{\text{d}x}\int_{0}^{x^2}\underbrace{\dfrac{\sin(t)}{t}}_{f(t)}\text{ d}t = \underbrace{\dfrac{\sin(x^2)}{x^2}}_{f(x^2)}\underbrace{(2x)}_{\text{deriv. of }x^2} - \dfrac{\text{d}}{\text{d}x}[F(0)] = \dfrac{2\sin(x^2)}{x}\text{.}$$

For a simpler and fundamental question, do you know what is $g'(x)$ when $$g(x)=\int^{x}_0 \frac{\sin(t)}{t}\ dt?$$ In general, do you know what is $g'(x)$ $$g(x)=\int_0^xh(t)\ dt?\tag{*}$$ Once you know what is $g'(x)$ for (*), can you find by chain rule what is the derivative of $f(x)=g(x^2)$?

The answer provided by @Clarinetist is the most efficient way to solve this problem. Here is another way.

Using the series definition of the sine function we have $$\sin(z) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} x^{2n-1}$$ and \begin{align} f(x) &= \int\limits_{0}^{x^{2}} \frac{\sin(z)}{z} \mathrm{d}z \\ &= \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} \int\limits_{0}^{x^{2}} z^{2n-2} \mathrm{d}z \end{align}

The integral on the right hand side above is $$\int\limits_{0}^{x^{2}} z^{2n-2} \mathrm{d}z = \frac{1}{2n-1} (x^{2})^{2n-1}$$ and now we have $$f(x) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!(2n-1)} (x^{2})^{2n-1}$$

Let $$g(x) = (x^{2})^{2n-1}$$ then $$\frac{\mathrm{d}g}{\mathrm{d}x} = \frac{2}{x}(2n-1)(x^{2})^{2n-1}$$

Now we have \begin{align} \frac{\mathrm{d}f}{\mathrm{d}x} &= \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!(2n-1)} \frac{\mathrm{d}g}{\mathrm{d}x} \\ &= \frac{2}{x} \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} (x^{2})^{2n-1} \\ &= \frac{2}{x} \sin(x^{2}) \end{align}