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Hi. Here's what I calculated for two integrals: P.V. \int_{-\infty}^{\infty}\frac{x}{(x-1)(x+1)}dx=\pi i-\pi i=0 and: P.V. \int_{-\infty}^{\infty}\frac{1}{(x-i)(x-1)(x+1)}dx=-\pi i+\pi i(1/2)=-\pi i/2

I think it depends on what f(x) is. For example, compute: P.V.\int_{-\infty}^{\infty} \frac{f(x)}{x^2-a^2}dx for: f(x)=x,x^2,x^3,\frac{1}{x-i}

I think it's quite an accomplishment to obtain a picture (from LiveScience) of a new viral infection. I can tell the virus has a set of "arms" attached all around it's outer coat. I assume this is for gaining entry into the cell membrane. Not sure though. Can someone explain what the purpose...

Hi. You need to plot it carefully and look at it: the lower boundary of the region is the function y=2/3x in one part of the region, but is the x-axis in the second part of the region. So . . . break it up into two integrals.

One more thing Random: If you really want to get good at these, learn to draw them. It then becomes crystal clear what's going on with the branches and also, it's quite a programming challenge to draw the complicated ones.

Ok then: \mathop\text{Res}_{z=-i} \frac{\sqrt{z}}{(z+i)(z-i)}=\frac{\sqrt{-i}}{-2i} and then calculate \sqrt{-i} using that argument.

You're calculating the residue at -i incorrectly for the branch you're using. Your branch is defined as: \sqrt{z}=r^{1/2}e^{i/2(\arg(z))},\quad 0\leq arg(z)<2\pi then the argument of -i is what?

Thanks for that information. It's interesting and I'll try to find a copy. :)

Draw a closed line integral and study it: Take a function of two variables f(x,y) which is nicely positive in the first quadrant. Now, draw a circle in the x-y plane in the first quadrant. Now, here it the tricky part. Draw the cylinder that connects the circle, straight up to the surface of...

"Chaos and Fractals" by Heinz-Otto Peitgen, et. al. and "An Introduction to Chaotic Dynamical Systems" by Robert Devaney.

Hello Caesius. May I make a suggestion I think would be helpful to you? Suppose all you had to do was to plot the surfaces. Never mind (for now) the integration. Could you do that, nicely? The surface z=0 is just the x-y plane right. The surface y=x^3 is a paraboloid sheet, and z=x is a...

See the thread below and how I solved it. Then apply the principles to your problem. https://www.physicsforums.com/showthread.php?t=310902&page=2

Use Reduce: f = Log[Erf[x/7]] - Cos[x^2 - 1] + 1; rts = Reduce[f == 0 && 1 < x < 6, x] myvals = N[x /. {ToRules[rts]}] I just used an example in Mathematica with a bunch of transcendental roots and then called Reduce to find all the zeros between 1 and 6. Reduce returns a list of...

Hi. Didn't notice you replied until now. << solution deleted by berkeman >>

Ok, I see what you mean. Here is a quote from the Wikipedia article: "NASA's original plan for safely de-orbiting Hubble was to retrieve it using a space shuttle. The Hubble telescope would then have most likely been displayed in the Smithsonian Institution. This is no longer considered...