To compute the integral of the function $\frac{\arctan\left(\frac{1}{t}\right)}{1 + t^2}$ from 0 to 1, we proceed with a substitution and careful analysis.

The integral is given by:

$I = \int_0^1 \frac{\arctan\left(\frac{1}{t}\right)}{1 + t^2} \, dt$

Step 1: Substitution

Let us make the substitution $t = \frac{1}{u}$. This transforms the variable $t$ in terms of $u$.

For $t = \frac{1}{u}$:

• $dt = -\frac{1}{u^2} du$
• When $t = 0$, $u \to \infty$
• When $t = 1$, $u = 1$

Thus, the limits of integration become from $\infty$ to 1, and the integral transforms as follows:

$I = \int_\infty^1 \frac{\arctan(u)}{1 + \frac{1}{u^2}} \cdot \left( -\frac{1}{u^2} \right) du$

Now simplify the denominator:

$1 + \frac{1}{u^2} = \frac{u^2 + 1}{u^2}$

Thus, the integral becomes:

$I = \int_1^\infty \frac{\arctan(u)}{u^2 + 1} \, du$

Step 2: Symmetry of the Integral

Now we see that the transformed integral looks very similar to the original one. In fact, it can be shown that:

$\int_0^1 \frac{\arctan\left(\frac{1}{t}\right)}{1 + t^2} \, dt = \int_1^\infty \frac{\arctan(t)}{1 + t^2} \, dt$

Thus, the original integral can be expressed as:

$I = \frac{1}{2} \int_0^\infty \frac{\arctan(t)}{1 + t^2} \, dt$

Step 3: Known Result

It is a known result that:

$\int_0^\infty \frac{\arctan(t)}{1 + t^2} \, dt = \frac{\pi}{2} \ln(2)$

Therefore:

$I = \frac{1}{2} \cdot \frac{\pi}{2} \ln(2) = \frac{\pi}{4} \ln(2)$

Final Answer:

The value of the integral is:

$I = \frac{\pi}{4} \ln(2)$

Would you like more details on any part of this process, or have any questions?

Here are 5 related questions to explore further:

1. How do you compute the integral of $\frac{\arctan(x)}{1 + x^2}$ from 0 to infinity?
2. What are other methods of integrating functions involving inverse trigonometric functions?
3. How does the substitution $t = \frac{1}{u}$ help simplify integrals?
4. Can the symmetry of functions always help reduce the complexity of definite integrals?
5. What are some applications of integrals involving $\arctan(x)$?

Tip: Always check for symmetries in an integral's limits or form—this can simplify the process significantly.