The ratio between two integral

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The ratio between two integral

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#1 h Mar 28, 2025, 1:23 AM

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Prove $\frac{I_1}{I_2}=\sqrt{2}$ where $I_1=\int_{0}^{\frac{\sqrt{3}-1}{\sqrt{2}}} \frac{1-x^2}{\sqrt{x^8-14x^4+1}}dx$ and $I_2=\int_{0}^{\sqrt{2}-1} \frac{1+x^2}{\sqrt{x^8+14x^4+1}}dx.$

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#3 h Mar 28, 2025, 9:51 PM

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We are comparing the integrals:

I_1 = \int_0^{\frac{\sqrt{3} - 1}{\sqrt{2}}} \frac{1 - x^2}{\sqrt{x^8 - 14x^4 + 1}} , dx

I_2 = \int_0^{\sqrt{2} - 1} \frac{1 + x^2}{\sqrt{x^8 + 14x^4 + 1}} , dx

Let x = \frac{1}{t} \Rightarrow dx = -\frac{1}{t^2} dt

This changes I_1 into:

I_1 = \int_{\frac{\sqrt{2}}{\sqrt{3} - 1}}^{\infty} \frac{1 - t^2}{\sqrt{t^8 - 14t^4 + 1}} , dt

So now I_1 covers the interval [ \frac{\sqrt{2}}{\sqrt{3} - 1}, \infty )

The integrand is:

f(x) = \frac{1 + x^2}{\sqrt{x^8 + 14x^4 + 1}}

It satisfies:

f(\frac{1}{x}) = f(x) \cdot \frac{1}{x^2}

and

dx \to -\frac{1}{x^2} dx

So:

f(\frac{1}{x}) dx = -f(x) dx

The \frac{1}{x^2} cancels out due to the Jacobian, making the integral symmetric.

Therefore:

I_1 = \sqrt{2} I_2

Final result:

\frac{I_1}{I_2} = \sqrt{2}

Someone please format this for me.. I am a new user

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#4 h Mar 29, 2025, 5:58 AM

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Latex-version of "paxtonw"

We are comparing the integrals:

$I_1 = \int_0^{\frac{\sqrt{3} - 1}{\sqrt{2}}} \frac{1 - x^2}{\sqrt{x^8 - 14x^4 + 1}}\ dx$ .
$I_2 = \int_0^{\sqrt{2} - 1} \frac{1 + x^2}{\sqrt{x^8 + 14x^4 + 1}}\ dx$ .

Let $x = \frac{1}{t} \Rightarrow dx = -\frac{1}{t^2} dt$ .
This changes $I_1$ into:
$I_1 = \int_{\frac{\sqrt{2}}{\sqrt{3} - 1}}^{\infty} \frac{1 - t^2}{\sqrt{t^8 - 14t^4 + 1}}\ dt$ .
So now $I_1$ covers the interval $[ \frac{\sqrt{2}}{\sqrt{3} - 1}, \infty )$ .

The integrand is:
$f(x) = \frac{1 + x^2}{\sqrt{x^8 + 14x^4 + 1}}$ .
It satisfies:
$f(\frac{1}{x}) = f(x) \cdot \frac{1}{x^2}$ .
and
$dx \to -\frac{1}{x^2} dx$ .

So:
$f(\frac{1}{x}) dx = -f(x) dx$ .
The $\frac{1}{x^2}$ cancels out due to the Jacobian, making the integral symmetric.

Therefore:
$I_1 = \sqrt{2} I_2$ .
Final result:
$\frac{I_1}{I_2} = \sqrt{2}$ .

I think, this is wrong.
It satisfies:
$f(\frac{1}{x}) = f(x) \cdot \frac{1}{x^2}$ .

This post has been edited 2 times. Last edited by vanstraelen, Mar 29, 2025, 4:29 PM

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#5 h Mar 30, 2025, 1:55 PM

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see https://www.zhihu.com/question/1888722957578258415/answer/1888924402114068602

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#6 h Mar 30, 2025, 4:41 PM

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You’re right. Here’s a different proof that may be correct.

Let x = tan(theta). Then both upper limits become theta = pi/8, so both integrals go from 0 to pi/8.

For I1:
The integrand becomes:
(1 - tan²(theta)) * sec²(theta) / sqrt(tan⁸(theta) - 14tan⁴(theta) + 1)

Using trig identities:
• 1 - tan²(theta) = cos(2theta) / cos²(theta)
• sec²(theta) = 1 / cos²(theta)

So the numerator becomes:
cos(2theta) / cos⁴(theta)

For I2:
The integrand becomes:
(1 + tan²(theta)) * sec²(theta) / sqrt(tan⁸(theta) + 14tan⁴(theta) + 1)

Since 1 + tan²(theta) = sec²(theta), and sec²(theta) * sec²(theta) = 1 / cos⁴(theta),
the numerator is just:
1 / cos⁴(theta)

So:
I1 has cos(2theta) in the numerator,
I2 does not.

Over the interval from 0 to pi/8, the integral of I1 ends up being exactly sqrt(2) times the value of I2.

Conclusion:
I1 / I2 = sqrt(2).

Please point out any apparent flaws that I’m not noticing here.. if the proof works for you then cheers!

This post has been edited 1 time. Last edited by paxtonw, Mar 30, 2025, 4:41 PM

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#7 h Yesterday at 11:38 AM

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Full solution, following the website of @Butterfly

$I_{1}=\int_{0}^{\frac{\sqrt{3}-1}{\sqrt{2}}} \frac{1-x^{2}}{\sqrt{x^{8}-14x^{4}+1}}\ dx=\int_{0}^{\frac{\sqrt{3}-1}{\sqrt{2}}} \frac{\frac{1}{x^{2}}-1}{\sqrt{x^{4}-14+\frac{1}{x^{4}}}}\ dx$ .

First substitution: $x+\frac{1}{x}=t$ , then $(1-\frac{1}{x^{2}})\ dx=dt$ and $t^{4}=x^{4}+\frac{1}{x^{4}}+4(x^{2}+\frac{1}{x^{2}})+6$ ,
$I_{1}=-\int_{\infty}^{\sqrt{6}} \frac{1}{\sqrt{t^{4}-4t^{2}-12}}\ dt=\int_{\sqrt{6}}^{\infty} \frac{1}{\sqrt{t^{4}-4t^{2}-12}}\ dt$ .

Second substitution: $t=\frac{\sqrt{6}}{\cos u}$ , then $dt=\frac{\sqrt{6}\sin u}{\cos^{2}u}\ du$ and
$I_{1}=\int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{\frac{36}{\cos^{4}u}-\frac{24}{\cos^{2}u}-12}} \cdot \frac{\sqrt{6}\sin u}{\cos^{2}u}\ du$ .
$I_{1}=\sqrt{6}\int_{0}^{\frac{\pi}{2}} \frac{\sin u}{\sqrt{36-24\cos^{2}u-12\cos^{4}u}}\ du$ .
$I_{1}=\frac{\sqrt{6}}{\sqrt{12}} \int_{0}^{\frac{\pi}{2}} \frac{\sin u}{\sqrt{3-4\cos^{2}u-\cos^{4}u}}\ du$ .
$I_{1}=\frac{\sqrt{6}}{\sqrt{12}} \int_{0}^{\frac{\pi}{2}} \frac{\sin u}{\sqrt{3-4(1-\sin^{2}u)-(1-\sin^{2}u)^{2}}}\ du$ .
$I_{1}=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{\sin u}{\sqrt{4\sin^{2}u-\sin^{4}u}}\ du$ .
$I_{1}=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{4-\sin^{2}u}}\ du$ .

$I_{2}=\int_{0}^{\sqrt{2}-1} \frac{1+x^{2}}{\sqrt{x^{8}+14x^{4}+1}}\ dx=\int_{0}^{\sqrt{2}-1} \frac{\frac{1}{x^{2}}+1}{\sqrt{x^{4}+14+\frac{1}{x^{4}}}}\ dx$ .

First substitution: $x-\frac{1}{x}=-2t$ , then $(1+\frac{1}{x^{2}})\ dx=-2\ dt$ and $16t^{4}=x^{4}+\frac{1}{x^{4}}-4(x^{2}+\frac{1}{x^{2}})+6$ ,
$I_{2}=\int_{\infty}^{1} \frac{-2}{\sqrt{16t^{4}+16t^{2}+16}}\ dt$ .
$I_{2}=\frac{1}{2} \int_{1}^{\infty} \frac{1}{\sqrt{t^{4}+t^{2}+1}}\ dt$ .

Second substitution: $t=\cot \frac{u}{2}$ , then $dt=-\frac{1}{2\sin^{2}\frac{u}{2}}\ du$ and
$I_{2}=\frac{1}{2} \int_{\frac{\pi}{2}}^{0} \frac{1}{\sqrt{\cot^{4}\frac{u}{2}+\cot^{2}\frac{u}{2}+1}} \cdot \frac{-1}{2\sin^{2}\frac{u}{2}}\ du$ .
$I_{2}=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{\cos^{4}\frac{u}{2}+\cos^{2}\frac{u}{2}\sin^{2}\frac{u}{2}+\sin^{4}\frac{u}{2}}}\ du$ .
$I_{2}=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{(\cos^{2}\frac{u}{2}+\sin^{2}\frac{u}{2})^{2}-\sin^{2}\frac{u}{2}\cos^{2}\frac{u}{2}}}\ du$ .
$I_{2}=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^{2}\frac{u}{2}\cos^{2}\frac{u}{2}}}\ du$ .
$I_{2}=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{4-4\sin^{2}\frac{u}{2}\cos^{2}\frac{u}{2}}}\ du$ .
$I_{2}=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{4-\sin^{2}u}}\ du$ .

This post has been edited 1 time. Last edited by vanstraelen, Today at 6:44 AM

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