# Given $f(x) = \int_{0}^{x^{20}}{\sqrt{t}\sin{\sqrt{t}}\,dt}$. Find $f^{(80)}(0)$

I want to find $$f^{(80)}(0)$$, where $$f(x) = \displaystyle \int_{0}^{x^{20}}{\sqrt{t}\sin{\sqrt{t}}\,dt}$$.

My attempt:

Let $$g(x) = \displaystyle \int{\sqrt{t}\sin{\sqrt{t}}\,dt}$$. Now $$f(x)$$ can be rewritten as $$f(x) = g{\left(x^{20}\right)} - g(0).$$

Taking the derivative $$80$$ times, $$f^{(80)}(x) = \sum_{i = 1}^{80}{a_i \times g^{(i)}{\left(x^{20}\right)}},$$ where $$\{a\}$$ is a sequence of integers. I tried to find the sum, but little progress was made.

I noticed that if I plugged $$x = 0$$ into the equation, all of the "$$x$$-linked" terms would be canceled. hence, only the first few terms would be significant, which is $$\displaystyle \sum_{i = 1}^{4}{a_i \times g^{(i)}{\left(x^{20}\right)}}$$.

Another observation is that $$a_i$$ will become $$0$$, i.e. $$g^{(i)}$$ will vanish after 20 times the derivative was taken. Thus, the final answer would be $$a_4 \times g^{(4)}{\left(x^{20}\right)}$$.

I only managed to find $$a_1 = 20!$$ for the first 20 derivatives and failed to derive explicit formulas for the other three.

I would like to know how to solve this problem appropriately, thanks in advance.

• Have you tried finding a closed form for $f(x)$? Substituting $u=\sqrt t$ and by-parts would help. Dec 19, 2020 at 18:05
• The function $f$ is not differentiable at the origin as it is not defined to the left of the origin. Dec 19, 2020 at 18:10
• @hunter Why do you think $f$ is not defined there? Actually, $f$ is an even function. Dec 19, 2020 at 18:22
• @hunter You might be confusing $f(x)$ with $f(t)$. Dec 19, 2020 at 18:26
• Think the power series way. Dec 19, 2020 at 18:27

Some simple rules allow us to find the first derivative

Namely the fundamental theorem of calculus and the chain rule

Which gives

$$\frac{df(x)}{dx} = \frac{df(x)}{d(x^{20})}\cdot \frac{d(x^{20})}{dx} = x^{10} \cdot \sin x^{10}\cdot 20x^{19}= 20x^{29}\sin x^{10}$$

We know that deriving this another 79 times gives the 80th derivative of $$f$$. To find the 79th derivative of this derivative we use

The taylor expansion of $$\sin u$$ at $$u=0$$ and substitute $$u=x^{10}$$

Which gives

$$20x^{29}\sin x^{10} = 20x^{29}\sum_{k=0}^{\infty} \frac{(-1)^k (x^{10})^{1 + 2 k}}{(1 + 2 k)!}$$

Which allows us to use the following fact

the 79th derivative of this function at zero will be $$79!$$ multiplied by the coefficient of $$x^{79}$$ - this is valid because all other terms have either vanished or are stil multiplied by $$x=0$$

We can find the corresponding $$k$$ as follows

$$10(1+2k) = 79-29 \iff k = 2$$

Now we apply arithmatic to find the solution

$$f^{(80)}(0) = \frac{20\cdot 79!}{5!} = \frac{79!}{6}$$

• Finding the derivative by Taylor's series is very new to me. Thank you for your detailed answer.
– KM02
Dec 19, 2020 at 20:03

We have $$f'(x) = 20x^{19}x^{10}\sin(x^{10}) = 20x^{29}\sin(x^{10})$$. What do the first few terms of the Taylor series of $$\sin$$ tell you?

• Thanks for the suggestion of using the Taylor series, I will try it.
– KM02
Dec 19, 2020 at 20:04

Let $$F(t) = \int{\sqrt{t}\sin{\sqrt{t}}\,dt}$$ Then, we have $$f(x) = F(x^{20})-F(0).$$

Differentiating once you get $$f'(x) = 20x^{19}F'(x^{20}) = 20x^{19}(\sqrt{x^{20}}\sin\sqrt{x^{20}}) \\ f'(x) = 20x^{29}\sin(x^{10})$$ I hope continuing is not difficult.

• Something's wrong. At the end you have $x^{19}\sqrt{x^{20}}=x^{39}$ Dec 19, 2020 at 18:09
• @saulspatz Oh yes, there should be an absolute value. Dec 19, 2020 at 18:17
• @VIVID No there shouldn't, it's something else :) Dec 19, 2020 at 18:18
• I see now, haha :D Dec 19, 2020 at 18:23