@Tanishq Varshney I don't think so this integration is possible with normal methods. Try substituting:
\[sinax=\frac{e^{iax}-e^{-iax}}{2i}\]
–
Aditya Kumar
·
1 year, 5 months ago

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I can integrate that, if \(a = 1\) or \(2\), and \(b = \dfrac{n\pi}{2} \). In that case, the answer to the integral would be \(= - \dfrac{n\pi}{2} \ln(2) \).

I don't think I can integrate it for other values of \(a, b \neq 0 \).
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Satyajit Mohanty
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1 year, 5 months ago

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It is a bit surprising that this can be integrated at all, since the function ln(sin(x)) becomes infinite as x -> 0. (But this can be proven by considering the integral of ln(x) from 0 to b.)

Although I doubt that the function ln(sin(ax)) can be integrated explicitly in terms of "elementary" functions — and we better assume that a and b are both positive here ((or else both negative. But we will assume they are both positive).

Now assume a = 1. We want the function ln(sin(x)) to be real, and for this we need to avoid values of sin(x) that are less than 0. So we must limit the values of x near 0 to 0 <= x <= pi.

So let us set F(x) := the integral from 0 to x of ln(sin(t)) dt.

We know that F(0) = 0. By using the advanced complex variables technique of contour integration (in an advanced way), it can be determined that F(pi/2) = -pi ln(2)/2 and, by the symmetry of sin(x), that also F(pi) = -pi ln(2).

If we like, we can extend this function to all x via F(x) := the integral from 0 to x of ln|sin(t)| dt.

Then the graph of y = F(x) is a beautiful wavy curve that wiggles about the line y = -ln(2) x, and takes the exact values of F(N pi/2) = -N pi ln(2)/2 for all integers N.

Extending these facts to values of a other than a = 1 is an easy task.
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Da Qu
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1 year, 5 months ago

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Let the integral be I. Now differentiate it with respect to a and after solving some more steps, you arrive at a differential equation in I and a i.e.,

dI/da + I/a = b*ln(sin(ab)).

Solve this differential equation to get I.

P.S.- Sorry, I dunno latex....else I would've posted my solution.
–
Appan Rakaraddi
·
1 year, 5 months ago

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TopNewest@Tanishq Varshney I don't think so this integration is possible with normal methods. Try substituting: \[sinax=\frac{e^{iax}-e^{-iax}}{2i}\] – Aditya Kumar · 1 year, 5 months ago

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I can integrate that, if \(a = 1\) or \(2\), and \(b = \dfrac{n\pi}{2} \). In that case, the answer to the integral would be \(= - \dfrac{n\pi}{2} \ln(2) \).

I don't think I can integrate it for other values of \(a, b \neq 0 \). – Satyajit Mohanty · 1 year, 5 months ago

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It is a bit surprising that this can be integrated at all, since the function ln(sin(x)) becomes infinite as x -> 0. (But this can be proven by considering the integral of ln(x) from 0 to b.)

Although I doubt that the function ln(sin(ax)) can be integrated explicitly in terms of "elementary" functions — and we better assume that a and b are both positive here ((or else both negative. But we will assume they are both positive).

Now assume a = 1. We want the function ln(sin(x)) to be real, and for this we need to avoid values of sin(x) that are less than 0. So we must limit the values of x near 0 to 0 <= x <= pi.

So let us set F(x) := the integral from 0 to x of ln(sin(t)) dt.

We know that F(0) = 0. By using the advanced complex variables technique of contour integration (in an advanced way), it can be determined that F(pi/2) = -pi ln(2)/2 and, by the symmetry of sin(x), that also F(pi) = -pi ln(2).

If we like, we can extend this function to all x via F(x) := the integral from 0 to x of ln|sin(t)| dt.

Then the graph of y = F(x) is a beautiful wavy curve that wiggles about the line y = -ln(2) x, and takes the exact values of F(N pi/2) = -N pi ln(2)/2 for all integers N.

Extending these facts to values of a other than a = 1 is an easy task. – Da Qu · 1 year, 5 months ago

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Let the integral be I. Now differentiate it with respect to a and after solving some more steps, you arrive at a differential equation in I and a i.e.,

dI/da + I/a = b*ln(sin(ab)).

Solve this differential equation to get I.

P.S.- Sorry, I dunno latex....else I would've posted my solution. – Appan Rakaraddi · 1 year, 5 months ago

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@Kartik Sharma @Isaac Buckley @Ishan Dasgupta Samarendra @Satyajit Mohanty – Tanishq Varshney · 1 year, 5 months ago

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