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determine F'(0)
EthanWYX2009   3
N an hour ago by Alphaamss
Source: 2024 Aug taca-13
Let
\[F(x)=\int\limits_0^{x}\left(\sin\frac 1t\right)^4\mathrm dt.\]Determine the value of $F'(0).$
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EthanWYX2009
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Alphaamss
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integral
Svyatoslav   5
N Nov 16, 2024 by tobiSALT
How do you find
$$I(a,b)=\int_0^\infty\frac{\ln|b-\cos ax|}{1+x^2}\,dx,\,\,b\in[0;1]\,\,?$$
5 replies
Svyatoslav
Nov 14, 2024
tobiSALT
Nov 16, 2024
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integral
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integration
calculus
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Svyatoslav
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Svyatoslav
#1 h  Nov 14, 2024, 6:05 AM • 1 Y
Y by Martin.s
How do you find
$$I(a,b)=\int_0^\infty\frac{\ln|b-\cos ax|}{1+x^2}\,dx,\,\,b\in[0;1]\,\,?$$
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tobiSALT
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tobiSALT
#2 h  Nov 15, 2024, 6:43 AM • 1 Y
Y by StarLex1
Let the integral be
$ I(a, b) = \int_0^{\infty} \frac{\ln |b - \cos(ax)|}{1 + x^2} \, dx, \quad b \in [0, 1] \text{ and } a > 0. $
Let $b = \cos \theta$ for some $\theta \in [0, \pi/2]$. Then we can rewrite the integral as
$ I(a, b) = \int_0^{\infty} \frac{\ln |\cos \theta - \cos(ax)|}{1 + x^2} \, dx. $
We have the identity $\cos \theta - \cos(ax) = 2 \sin \frac{ax + \theta}{2} \sin \frac{ax - \theta}{2}$. Thus,
$ \ln |\cos \theta - \cos(ax)| = \ln |2 \sin \frac{ax+\theta}{2} \sin \frac{ax-\theta}{2}| = \ln 2 + \ln |\sin \frac{ax+\theta}{2}| + \ln |\sin \frac{ax-\theta}{2}|. $
Also, we know that
$ \int_0^{\infty} \frac{\ln 2}{1 + x^2} \, dx = \ln 2 \int_0^{\infty} \frac{dx}{1 + x^2} = \ln 2 \arctan(x) \Big|_0^{\infty} = \ln 2 \cdot \frac{\pi}{2} = \frac{\pi}{2} \ln 2. $
We use the known result
$ \int_0^{\infty} \frac{\ln|\sin(rx)|}{1 + x^2} \, dx = \frac{\pi}{2} \ln \left( \frac{1 - e^{-2r}}{2} \right). $
Then
$ \int_0^{\infty} \frac{\ln |\sin \frac{ax+\theta}{2}|}{1 + x^2} \, dx = \int_0^{\infty} \frac{\ln |\sin \frac{a}{2} (x+\theta/a)|}{1 + x^2} \, dx $
and
$ \int_0^{\infty} \frac{\ln |\sin \frac{ax-\theta}{2}|}{1 + x^2} \, dx = \int_0^{\infty} \frac{\ln |\sin \frac{a}{2} (x-\theta/a)|}{1 + x^2} \, dx. $
So we have
$ I(a, b) = \int_0^{\infty} \frac{\ln |b - \cos(ax)|}{1 + x^2} \, dx = \int_0^{\infty} \frac{\ln |\cos \theta - \cos(ax)|}{1 + x^2} \, dx = \frac{\pi}{2} \ln 2 + \int_0^{\infty} \frac{\ln |\sin(\frac{ax+\theta}{2})| + \ln |\sin(\frac{ax-\theta}{2})|}{1 + x^2} \, dx. $
If $\theta = 0$, i.e., $b = 1$, then
$ \int_0^{\infty} \frac{\ln(1 - \cos(ax))}{1 + x^2} \, dx = \int_0^{\infty} \frac{2 \ln |\sin(ax/2)| + \ln 2}{1 + x^2} \, dx = \frac{\pi}{2} \ln 2 + 2 \int_0^{\infty} \frac{\ln |\sin(ax/2)|}{1 + x^2} \, dx. $
Using the formula above, we get
$ \frac{\pi}{2} \ln 2 + 2 \frac{\pi}{2} \ln \left( \frac{1 - e^{-a}}{2} \right) = \frac{\pi}{2} (\ln 2 + 2 \ln (1 - e^{-a}) - 2 \ln 2) = \pi \ln(1 - e^{-a}) - \frac{\pi}{2} \ln 2. $
Thus
$ I(a, 1) = \pi \ln(1 - e^{-a}) - \frac{\pi}{2} \ln 2. $
Therefore, our final answer is
$ I(a, b) = \pi \ln \left( \frac{1 + \sqrt{1-b^2} - e^{-a}}{2} \right). $
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Svyatoslav
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Svyatoslav
#3 h  Nov 15, 2024, 12:54 PM • 1 Y
Y by tobiSALT
Thank you for your solution, but I think the answer is not correct. At $b=0\,$ and $\,a=0$
$$\int_0^\infty\frac{\ln (1)}{1+x^2}dx=0\neq\,\pi\ln\frac12$$There is also a known result (I.S. Gradshteyn and I.M. Ryzhik, GW (338)(28a), p 569)
$$\int_0^\infty\frac{\ln\cos^2ax}{1+x^2}dx=\pi\ln\frac{1+e^{-2a}}2$$
This post has been edited 1 time. Last edited by Svyatoslav, Nov 15, 2024, 6:42 PM
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tobiSALT
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tobiSALT
#4 h  Nov 15, 2024, 1:08 PM
Y by
True. I´ll try again later today, this integral is really hard.
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Svyatoslav
537 posts
Svyatoslav
#5 h  Nov 15, 2024, 1:20 PM • 1 Y
Y by Martin.s
Thank you. I got
$$I(a,b)=\frac\pi2\ln\frac{1-2be^{-a}+e^{-2a}}2, \,b\in [-1;1];\,a\geqslant0$$what gives the correct answer for one of the options: $b=0$ or $b=1$ or $a=0$. My problem is that WA (free option) is not able to evaluate the integral numerically, and I cannot get a reliable numeric check.

PS The answer is indeed correct - please see the solution.
This post has been edited 2 times. Last edited by Svyatoslav, Nov 16, 2024, 1:18 AM
Reason: slight correction
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tobiSALT
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tobiSALT
#7 h  Nov 16, 2024, 5:07 PM • 1 Y
Y by Svyatoslav
Great! Good solution
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