Hello everyone,

I'm trying to reproduce the example integral on MathWorld for Leibniz
Integral Rule

http://mathworld.wolfram.com/LeibnizIntegralRule.html

namely,

\phi[\alpha] = \int_0^\pi log[1 - 2 \alpha \cos[x] + \alpha^2 ] dx = 2
\pi log | \alpha |

It appears a little strange that, the LHS and RHS do not match up when
\alpha \to 0,

since the former would be int_0^\pi log[1] dx = 0,
whereas the latter would be 2 \pi log[0] = -\infty. Am I missing
something?

I actually got up to

\phi[\alpha] = 2 \pi log | \alpha | + constant

by differentiating w.r.t \alpha, changing variables to t = \tan[x/2],
obtaining an even integrand in t, extending the limits from {0, \infty}
to {-\infty, \infty} and then using contour integration. But I'm unable
to show that the above constant is zero as asserted in MathWorld.

Thanks for the help, and happy holidays!

Integrate that x=0 to Pi to get 2*pi/a, here we must assume a>2.

In article <1166845444.134113.318100@i12g2000cwa.googlegroups.com>,
wandering.the.cosmos@gmail.com> wrote:

Hello everyone,

I'm trying to reproduce the example integral on MathWorld for Leibniz
Integral Rule

http://mathworld.wolfram.com/LeibnizIntegralRule.html

namely,

\phi[\alpha] = \int_0^\pi log[1 - 2 \alpha \cos[x] + \alpha^2 ] dx = 2
\pi log | \alpha |

It appears a little strange that, the LHS and RHS do not match up when
\alpha \to 0,

Note the condition |alpha| > 1 ?? When -1 < alpha < 1, then the
integrand is unbounded, and there are negative values inside the
logarithm.

= 0 for all real 'a' and real 'x'

since the former would be int_0^\pi log[1] dx = 0,
whereas the latter would be 2 \pi log[0] = -\infty. Am I missing
something?

I actually got up to

\phi[\alpha] = 2 \pi log | \alpha | + constant

by differentiating w.r.t \alpha, changing variables to t = \tan[x/2],
obtaining an even integrand in t, extending the limits from {0, \infty}
to {-\infty, \infty} and then using contour integration. But I'm unable
to show that the above constant is zero as asserted in MathWorld.

Thanks for the help, and happy holidays!


Take the integrand log(1-2*a*cos(x)+a^2)
Differentiate with respect to a, to get
(-2*cos(x)+2*a)/(1-2*a*cos(x)+a^2)
Integrate that x=0 to Pi to get 2*pi/a, here we must assume a>2.
Integrate with respect to a to get the answer.

As you know, putting a=0 to get the constant doesn't
work. Try letting a -> infinity to do that.
The original integral is asymptotic to 2*Pi*log(a), so
the constant must be zero.

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/

As you know, putting a=0 to get the constant doesn't
work. Try letting a -> infinity to do that.
The original integral is asymptotic to 2*Pi*log(a), so
the constant must be zero.

Hello everyone,

I'm trying to reproduce the example integral on MathWorld for Leibniz
Integral Rule

http://mathworld.wolfram.com/LeibnizIntegralRule.html

namely,

\phi[\alpha] = \int_0^\pi log[1 - 2 \alpha \cos[x] + \alpha^2 ] dx = 2
\pi log | \alpha |

It appears a little strange that, the LHS and RHS do not match up when
\alpha \to 0, since the former would be int_0^\pi log[1] dx = 0,
whereas the latter would be 2 \pi log[0] = -\infty. Am I missing
something?

I actually got up to

\phi[\alpha] = 2 \pi log | \alpha | + constant

by differentiating w.r.t \alpha, changing variables to t = \tan[x/2],
obtaining an even integrand in t, extending the limits from {0, \infty}
to {-\infty, \infty} and then using contour integration. But I'm unable
to show that the above constant is zero as asserted in MathWorld.

Thanks for the help, and happy holidays!

In article <1166845444.134113.318100@i12g2000cwa.googlegroups.com>,
wandering.the.cosmos@gmail.com> wrote:

Hello everyone,

I'm trying to reproduce the example integral on MathWorld for Leibniz
Integral Rule

http://mathworld.wolfram.com/LeibnizIntegralRule.html

namely,

\phi[\alpha] = \int_0^\pi log[1 - 2 \alpha \cos[x] + \alpha^2 ] dx = 2
\pi log | \alpha |

It appears a little strange that, the LHS and RHS do not match up when
\alpha \to 0,

Note the condition |alpha| > 1 ?? When -1 < alpha < 1, then
the integrand is unbounded, and there are negative values inside the
logarithm.

since the former would be int_0^\pi log[1] dx = 0,
whereas the latter would be 2 \pi log[0] = -\infty. Am I missing
something?

I actually got up to

\phi[\alpha] = 2 \pi log | \alpha | + constant

by differentiating w.r.t \alpha, changing variables to t = \tan[x/2],
obtaining an even integrand in t, extending the limits from {0, \infty}
to {-\infty, \infty} and then using contour integration. But I'm unable
to show that the above constant is zero as asserted in MathWorld.

Thanks for the help, and happy holidays!


Take the integrand log(1-2*a*cos(x)+a^2)
Differentiate with respect to a, to get
(-2*cos(x)+2*a)/(1-2*a*cos(x)+a^2)
Integrate that x=0 to Pi to get 2*pi/a, here we must assume a>2.
Integrate with respect to a to get the answer.

As you know, putting a=0 to get the constant doesn't
work. Try letting a -> infinity to do that.
The original integral is asymptotic to 2*Pi*log(a), so
the constant must be zero.

If |a| < 1, then the integral of (-2*cos(x)+2*a)/(1-2*a*cos(x)+a^2) is
0,

On Jan 4, 9:05 pm, Waldek Hebisch <hebi...@math.uni.wroc.pl> wrote:

If |a| < 1, then the integral of (-2*cos(x)+2*a)/(1-2*a*cos(x)+a^2) is
0,

Could you explain why the integral is zero for |a|<1?

Thanks!

Cauchy's integral theorem. The integral of dz/(z-1) around the circle
|z|=a is zero if the pole 1 is outside the circle.