Let $$ I = \int_0^1 \frac{1}{1+t} dt + \frac{\pi i}{2} \int_0^1 \frac{1}{1+e^2} dt - i \int_0^1 \frac{1}{1+it} dt, $$ where $ t $ is a real variable and $ i = \sqrt{-1} $. The value of $ I $ is ______________
The problem asks for the value of the complex integral $ I $ defined as:
$ I = \int_0^1 \frac{1}{1+t} dt + \frac{\pi i}{2} \int_0^1 \frac{1}{1+e^2} dt - i \int_0^1 \frac{1}{1+it} dt $We calculate each term separately.
Term 1: Real Integral
The first term is a standard definite integral:
$ \int_0^1 \frac{1}{1+t} dt $The antiderivative of $ \frac{1}{1+t} $ is $ \ln(1+t) $. Evaluating the definite integral:
$ [\ln(1+t)]_0^1 = \ln(1+1) - \ln(1+0) = \ln(2) - \ln(1) = \ln(2) $Term 2: Constant Integral with Complex Factor
The second term involves an integral of a constant with respect to $ t $:
$ \int_0^1 \frac{1}{1+e^2} dt $Since $ \frac{1}{1+e^2} $ is a constant:
$ \int_0^1 \frac{1}{1+e^2} dt = \frac{1}{1+e^2} \int_0^1 dt = \frac{1}{1+e^2} [t]_0^1 = \frac{1}{1+e^2} (1 - 0) = \frac{1}{1+e^2} $Multiplying by the factor $ \frac{\pi i}{2} $ gives the value of the second part of $ I $:
$ \frac{\pi i}{2} \cdot \frac{1}{1+e^2} = \frac{\pi i}{2(1+e^2)} $Term 3: Complex Integral
The third term is:
$ - i \int_0^1 \frac{1}{1+it} dt $To evaluate $ \int_0^1 \frac{1}{1+it} dt $, we can use a substitution $ u = 1+it $. Then $ du = i dt $, so $ dt = \frac{du}{i} $. The limits of integration change from $ t=0 \to u=1 $ and $ t=1 \to u=1+i $.
$ \int_0^1 \frac{1}{1+it} dt = \int_1^{1+i} \frac{1}{u} \left( \frac{du}{i} \right) = \frac{1}{i} [\ln u]_1^{1+i} $Calculate $ \ln(1+i) $. In polar form, $ 1+i = \sqrt{2} e^{i\pi/4} $. Therefore, $ \ln(1+i) = \ln(\sqrt{2}) + i \frac{\pi}{4} = \frac{1}{2}\ln 2 + i \frac{\pi}{4} $.
So, the integral becomes:
$ \frac{1}{i} \left( \frac{1}{2}\ln 2 + i \frac{\pi}{4} - \ln(1) \right) = \frac{1}{i} \left( \frac{1}{2}\ln 2 + i \frac{\pi}{4} \right) = \frac{\frac{1}{2}\ln 2}{i} + \frac{i \frac{\pi}{4}}{i} = -i \frac{1}{2}\ln 2 + \frac{\pi}{4} $The third term of $ I $ is therefore:
$ -i \left( \frac{\pi}{4} - i \frac{\ln 2}{2} \right) = -i \frac{\pi}{4} + i^2 \frac{\ln 2}{2} = -i \frac{\pi}{4} - \frac{\ln 2}{2} $Combining the Terms
Now, substitute the evaluated terms back into the expression for $ I $:
$ I = \ln(2) + \frac{\pi i}{2(1+e^2)} + \left( -\frac{\ln 2}{2} - i \frac{\pi}{4} \right) $Group the real and imaginary parts:
$ I = \left( \ln 2 - \frac{\ln 2}{2} \right) + i \left( \frac{\pi}{2(1+e^2)} - \frac{\pi}{4} \right) $Simplify the expression:
$ I = \frac{\ln 2}{2} + i \frac{\pi}{4} \left( \frac{2}{1+e^2} - 1 \right) $ $ I = \frac{\ln 2}{2} + i \frac{\pi}{4} \left( \frac{2 - (1+e^2)}{1+e^2} \right) $ $ I = \frac{\ln 2}{2} + i \frac{\pi (1 - e^2)}{4 (1+e^2)} $This is the final simplified expression for the complex integral $ I $.
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