Prove that $\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}~dx=0$ (without trigonometric substitution)

Question

The integral is from P. Nahin's "Inside Interesting Integrals...", problem C2.1. His proposed solution includes trigonometric substitution and the use of log-sine integral.

However, I think the problem should have an easier solution (without appealing to another complicated integral at least).

I have the following trick in mind. Let's introduce the substitution $x=4-z$

$$I=\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}~dx=\int_0^4 \frac{\ln (4-z)}{\sqrt{4z-z^2}}~d(4-z)=\int_0^4 \frac{\ln (4-z)}{\sqrt{4z-z^2}}~dz$$

$$2I=\int_0^4 \frac{\ln (4x-x^2)}{\sqrt{4x-x^2}}~dx$$

$$I=\int_0^4 \frac{\ln \sqrt{4x-x^2}}{\sqrt{4x-x^2}}~dx$$

---

And here I'm stuck. I'm not sure if this can go somewhere. Maybe partial integration can help, but I don't know how to choose the functions. What do you think?

---

Here is a question about this integral.

Only one answer does not use trig substitution, it used gamma function instead. If there are no other ways, I'm prepared to give up on my question. But I would be grateful if it's left open at least for several days

---

Edit

After many attempts, I conclude that there is no trick to this integral. The reason is: the general form of this integral in not zero, but has the same symmetry properties, as the above case:

$$I(a)=\int_0^a \frac{\ln x}{\sqrt{ax-x^2}}~dx=\int_0^a \frac{\ln (a-x)}{\sqrt{ax-x^2}}~dx=\int_0^a \frac{\ln \sqrt{ax-x^2}}{\sqrt{ax-x^2}}~dx \neq 0$$

$$I(4)=0$$

So we will get nothing from symmetry considerations alone. There are two possible ways to solve this - either trigonometric substitution or gamma function.

Edit 2

I was wrong it seems, see the accepted answer.

Answer 1

Notice that by the substitution $x = 2 + u$,

$$ I = \int_{-2}^{2} \frac{\log(2 + u)}{\sqrt{4 - u^2}} \, du = \int_{0}^{2} \frac{\log(4 - u^2)}{\sqrt{4 - u^2}} \, du. $$

On the other hand, by the substitution $x = 4 - v^2$ (or equivalently $v = \sqrt{4 - x}$), we have

$$ I = \int_{0}^{2} \frac{\log(4 - v^2)}{v \sqrt{4 - v^2}} \cdot 2v \, dv = 2 \int_{0}^{2} \frac{\log(4 - v^2)}{\sqrt{4 - v^2}} \, dv. $$

Comparing two formulas give $I = 2I$ and therefore $I = 0$.

Answer 2

$\displaystyle K=\int_0^4 \dfrac{\ln (4x-x^2)}{\sqrt{4x-x^2}}~dx$

$\displaystyle K=\int_0^4 \dfrac{\ln (4-x)}{\sqrt{4x-x^2}}~dx+\int_0^4 \dfrac{\ln x}{\sqrt{4x-x^2}}~dx$

By the change of variable $y=4-x$, it's readily seen that the two preceding integrals are equal.

$\displaystyle K=\int_0^2 \dfrac{\ln (4x-x^2)}{\sqrt{4x-x^2}}~dx+\int_2^4 \dfrac{\ln (4x-x^2)}{\sqrt{4x-x^2}}~dx$

Perform the change of variable $y=\sqrt{4x-x^2}$ in both preceding integrals,

$\displaystyle K=2\int_0^2 \dfrac{\ln x}{\sqrt{4-x^2}}dx+2\int_0^2 \dfrac{\ln x}{\sqrt{4-x^2}}dx=4\int_0^2 \dfrac{\ln x}{\sqrt{4-x^2}}dx$

Thus,

$\displaystyle K=2\int_0^2 \dfrac{\ln x}{\sqrt{1-\left(\tfrac{x}{2}\right)^2}}dx$

Perform the change of variable $y=\dfrac{x}{2}$,

$\displaystyle K=4\int_0^1 \dfrac{\ln(2x)}{\sqrt{1-x^2}}dx$

$\displaystyle K=4\int_0^1 \dfrac{\ln 2}{\sqrt{1-x^2}}dx+4\int_0^1 \dfrac{\ln x}{\sqrt{1-x^2}}dx$

Perform the change of variable $x=\sin y$ in both integrals,

$\displaystyle K=4\ln(2) \int_0^{\tfrac{\pi}{2}} \dfrac{\cos y}{\sqrt{1-(\sin y)^2}}dy+4\int_0^{\tfrac{\pi}{2}} \dfrac{\ln (\sin y)\cos y}{\sqrt{1-(\sin y)^2}}dy=2\pi\ln 2+4\int_0^{\tfrac{\pi}{2}} \ln (\sin y)dy$

It is well known that $\displaystyle \int_0^{\tfrac{\pi}{2}} \ln (\sin y)dy=-\dfrac{\pi \log 2}{2}$

Thus, $K=0.$

Finally we get:

$\displaystyle \int_0^4 \dfrac{\ln (4-x)}{\sqrt{4x-x^2}}~dx=\int_0^4 \dfrac{\ln x}{\sqrt{4x-x^2}}~dx=0$

PS: To be compliant with the question.

$\displaystyle \int_0^1 \dfrac{1}{\sqrt{1-x^2}}dx=\dfrac{1}{2}\int_0^1 x^{\tfrac{1}{2}-1}(1-x)^{\tfrac{1}{2}-1}dx=\dfrac{\Gamma\left(\tfrac{1}{2}\right)^2}{\Gamma(1)}=\dfrac{\pi}{2}$

$\displaystyle \int_0^1 \dfrac{\ln x}{\sqrt{1-x^2}}dx=\dfrac{1}{4}\dfrac{\partial}{\partial s}\left[\int_0^1 x^s(1-x)^{\tfrac{1}{2}-1}dx\right]_{s=-\tfrac{1}{2}}=\dfrac{1}{4}\dfrac{\partial}{\partial s}\left[\dfrac{\Gamma(s+1)\Gamma\left(\tfrac{1}{2}\right)}{\Gamma\left(s+1+\tfrac{1}{2}\right)}\right]_{s=-\tfrac{1}{2}}=-\dfrac{\pi\ln2}{2}$

Answer 3

sketch: put $\sqrt{4x-x^2}=\sqrt{x(4-x)}=xr(t)$,$r(t)$ to be determined. On squaring we get $4-x=xr^2(t)$, thereby $x$ is a rational function of $t$ when $r(t)=\sqrt{t}$. This way we have got rid of the radical. The rest can be dealt with through inrtegration by parts.

Answer 4

I want to provide a generalization of @SangchulLee method for the integral:

$$I(a)=\int_0^a \frac{\ln x}{\sqrt{ax-x^2}}~dx=\int_0^a \frac{\ln (a-x)}{\sqrt{ax-x^2}}~dx=\int_0^a \frac{\ln \sqrt{ax-x^2}}{\sqrt{ax-x^2}}~dx$$

Let's make a change of variable:

$$x=\frac{a}{2}+u~~~~~~~a-x=\frac{a}{2}-u$$

$$I(a)=\frac{1}{2} \int_{-a/2}^{a/2} \frac{\ln (\frac{a^2}{4}-u^2)}{\sqrt{\frac{a^2}{4}-u^2}}~du=\int_{0}^{a/2} \frac{\ln (\frac{a^2}{4}-u^2)}{\sqrt{\frac{a^2}{4}-u^2}}~du$$

Let's make another change of variable:

$$x=a-v^2~~~~~~~a-x=v^2$$

$$I(a)=2 \int_{0}^{\sqrt{a}} \frac{\ln (a-v^2)}{\sqrt{a-v^2}}~dv$$

---

$$I(4)=2 I(4),~~~~~~I(4)=0 $$

However, we can make even more general conclusion. Let's denote $b=a/2$ and:

$$J(b)=\int_{0}^{b} \frac{\ln (b^2-t^2)}{\sqrt{b^2-t^2}}~dt$$

$$J(b)=2 J(\sqrt{2 b})$$

I already know the answer of course, but it it possible to guess:

$$J(b)=C_1 \ln (C_2 b)$$

$$C_1 \ln (C_2 b)=2 C_1 \ln (C_2 \sqrt{2 b})$$

$$C_1 \ln (C_2)+C_1 \ln (b)=2 C_1 \ln (C_2)+C_1 \ln (2)+C_1 \ln (b)$$

$$C_2=\frac{1}{2}$$

$$J(b)=C_1 \ln \left( \frac{b}{2} \right)$$

Which is correct and we only need to find one constant to complete the solution. Which of course requires solving the integral the 'honest' way.

Answer 5

\begin{align}J&=\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}~dx\\ &\overset{u=4-x}=\int_0^4 \frac{\ln(4-u)}{\sqrt{u(4-u}}~du\\ 2J&\overset{z=\sqrt{\frac{u}{4-u}}}=2\int_0^\infty \frac{\ln\left(\frac{16z^2}{(1+z^2)^2}\right)}{1+z^2}dz\\ &=8\ln 2 \underbrace{\int_0^\infty\frac{1}{1+z^2}dz}_{=\frac{\pi}{2}}+4\underbrace{\int_0^\infty \frac{\ln z}{1+z^2}dz}_{=0}-4\underbrace{\int_0^\infty \frac{\ln(1+z^2)}{1+z^2}dz}_{z=\tan t}\\ &=4\ln 2+8\underbrace{\int_0^{\frac{\pi}{2}}\ln(\cos t)dt}_{=-\frac{\pi\ln 2}{2}}=4\ln 2-4\ln 2=0 \end{align}

PS: The first change of variable is useless. Direct change of variable $u=\sqrt{\frac{x}{4-x}}$ works fine.