MHB Complex Integral: $\int_0^1$ Calculation

[Dustinsfl]

$\displaystyle \int_0^1 \frac{2t+i}{t^2+it+1} dt = \int_0^1 \left(\frac{t}{2} + \frac{i}{4} + \frac{5/4}{2t+i}\right) dt = \frac{1}{4} + \frac{5}{8} \ln\left(\sqrt{5}\right) + i\left(\frac{1}{4} + \frac{5}{8}\tan^{-1}\left(\frac{1}{2}\right)\right)$

Is this correct?

 

[Ackbach]

$\displaystyle \int_0^1 \frac{2t+i}{t^2+it+1} dt = \int_0^1 \left(\frac{t}{2} + \frac{i}{4} + \frac{5/4}{2t+i}\right) dt = \frac{1}{4} + \frac{5}{8} \ln\left(\sqrt{5}\right) + i\left(\frac{1}{4} + \tan^{-1}(-2)\right)$

Is this correct?

I don't think it is. Note that the numerator is the derivative of the denominator. What does that suggest?

 

[Dustinsfl]

I don't think it is. Note that the numerator is the derivative of the denominator. What does that suggest?

U-sub isn't defined for Complex integrals because any closed path would be zero. A counter example is 1/z around the unit circle which isn't 0 and is closed path.

 

[polygamma]

What Ackbach suggests is correct because you're integrating with respect to a real variable ($t$), in which case $i$ is just a constant.

 

[Ackbach]

U-sub isn't defined for Complex integrals because any closed path would be zero. A counter example is 1/z around the unit circle which isn't 0 and is closed path.

What Ackbach suggests is correct because you're integrating with respect to a real variable ($t$), in which case $i$ is just a constant.

Also, the denominator is nonzero on the integration path.

 

[Dustinsfl]

What Ackbach suggests is correct because you're integrating with respect to a real variable ($t$), in which case $i$ is just a constant.

I think I understand what my professor means. If we substitute,

$\int_{\gamma}\frac{1}{z}dz = \int_0^{2\pi}\frac{ie^{it}}{e^{it}}dt$

The numerator is the derivative of the denominator so substitutions is viable here as well.

$\int_1^1\frac{du}{u}=0\neq 2\pi i$.

 

[polygamma]

EDIT: Erased a bunch of nonsesne

 

[polygamma]

Ignore my previous post. It contains a bit of nonsense.$u = e^{it}$ for $t$ from $0$t o $2 \pi$ is the unit circle.

So $\displaystyle \int_{\gamma}\frac{1}{z}dz = \int_0^{2\pi}\frac{ie^{it}}{e^{it}}dt {\color {red} \ne } \int_{1}^{1}\frac{1}{u}du$.

But rather $\displaystyle \int_{\gamma}\frac{1}{z}dz = \int_0^{2\pi}\frac{ie^{it}}{e^{it}}dt = \int_{\gamma}\frac{1}{u}du$. And we're backing to where we started.

 

[Dustinsfl]

Ignore my previous post. It contains a bit of nonsense.$u = e^{it}$ for $t$ from $0$t o $2 \pi$ is the unit circle.

So $\displaystyle \int_{\gamma}\frac{1}{z}dz = \int_0^{2\pi}\frac{ie^{it}}{e^{it}}dt {\color {red} \ne } \int_{1}^{1}\frac{1}{u}du$.

But rather $\displaystyle \int_{\gamma}\frac{1}{z}dz = \int_0^{2\pi}\frac{ie^{it}}{e^{it}}dt = \int_{\gamma}\frac{1}{u}du$. And we're backing to where we started.

I just multiplied by the conjugate and obtained the answer without the use of substitution.

 

[polygamma]

So you said that $\displaystyle \int_{0}^{1} \frac{2t +i}{t^{2}+it+1} \ dt \ \frac{t^{2}-it+1}{t^2-it+1} = \int_{0}^{1} \frac{2t^{3}+3t}{t^4+3t^{2}+1} \ dt + i \int_{0}^{1} \frac{1-t^{2}}{t^{4}+3t^{2}+1} \ dt $?

The first integral is easy, but the second integral looks fairly nasty. But that is a valid approach.But I would still make that substitution to get $\displaystyle \int \frac{2t +i}{t^{2}+it+1} \ dt = \int \frac{1}{u} \ du = \ln u + C = \ln(t^{2}+it+1) + C$

Then $\displaystyle \int_{0}^{1} \frac{2t +i}{t^{2}+it+1} \ dt = \ln(1^{2}+i(1)+1) - \ln(0^{2}+i(0)+1) = \ln(2+i) - \ln(1) = \ln(2+i)$ which is the answer that WolframAlpha gives

http://www.wolframalpha.com/input/?i=integrate+(2t%2Bi)%2F(t^2%2Bit%2B1)+from+0+to+1

 

[Dustinsfl]

So you said that $\displaystyle \int_{0}^{1} \frac{2t +i}{t^{2}+it+1} \ dt \ \frac{t^{2}-it+1}{t^2-it+1} = \int_{0}^{1} \frac{2t^{3}+3t}{t^4+3t^{2}+1} \ dt + i \int_{0}^{1} \frac{1-t^{2}}{t^{4}+3t^{2}+1} \ dt $?

The first integral is easy, but the second integral looks fairly nasty. But that is a valid approach.But I would still make that substitution to get $\displaystyle \int \frac{2t +i}{t^{2}+it+1} \ dt = \int \frac{1}{u} \ du = \ln u + C = \ln(t^{2}+it+1) + C$

Then $\displaystyle \int_{0}^{1} \frac{2t +i}{t^{2}+it+1} \ dt = \ln(1^{2}+i(1)+1) - \ln(0^{2}+i(0)+1) = \ln(2+i) - \ln(1) = \ln(2+i)$ which is the answer that WolframAlpha gives

http://www.wolframalpha.com/input/?i=integrate+(2t+i)/(t^2+it+1)+from+0+to+1

Yup that is what I did.
Since u-sub isn't defined for complex integrals, we can't use it even though it may work in some cases.

 

[polygamma]

Can you quote where in your textbook such a claim is made?

 

[Dustinsfl]

Can you quote where in your textbook such a claim is made?

It is made by Richard Foote.

 

[polygamma]

I do believe you misinterpreted what he said. Because to evaluate every complex integral by breaking it into it's real and imaginary parts will become ridiculously time-consuming.

 

[Dustinsfl]

I do believe you misinterpreted what he said. Because to evaluate every complex integral by breaking it into it's real and imaginary parts will become ridiculously time-consuming.

I don't do that every time. He wanted us to appreciate Cauchy's Integral Formula, Residue Theory, etc. So we were doing these integrals the hard way. I didn't misinterpret. He had a had a brief discussion about it on Friday when he noticed that some students were using it. That is when he gave the counter example of all closed curves will evaluate to 0 when we know that isn't the case.

 

[polygamma]

But your assertion that making the substitution $u =e^{it}$ would mean that $\displaystyle \int_{0}^{2 \pi} \frac{i e^{it}}{e^{it}} \ dt = \int_{1}^{1} \frac{1}{u} \ du$ is false.

That substitution, as I already stated, would mean that $\displaystyle \int_{0}^{2 \pi} \frac{i e^{it}}{e^{it}} \ dt = \int_{\gamma} \frac{1}{u} \ du$

 

[Dustinsfl]

But your assertion that making the substitution $u =e^{it}$ would mean that $\displaystyle \int_{0}^{2 \pi} \frac{i e^{it}}{e^{it}} \ dt = \int_{1}^{1} \frac{1}{u} \ du$ is false.

That substitution, as I already stated, would mean that $\displaystyle \int_{0}^{2 \pi} \frac{i e^{it}}{e^{it}} \ dt = \int_{\gamma} \frac{1}{u} \ du$

If you do a change of bounds, you get 1 and 1. e^0 = 1 and e^{2\pi i} = 1

 

[polygamma]

We made the substitution $u = e^{it}$, and the limits are from $t=0$ to $t= 2 \pi$. Isn't that the definition of the unit circle?