To solve the integral

$\int_0^\infty \frac{dx}{1+x^4}$

we can use the following approach:

Step 1: Substitution to Simplify the Integral

Let $x^2 = t$, which implies $x = \sqrt{t}$ and $dx = \frac{1}{2\sqrt{t}} dt$.

• When $x = 0$, $t = 0$.
• When $x \to \infty$, $t \to \infty$.

Substituting into the integral, we have: $\int_0^\infty \frac{dx}{1+x^4} = \int_0^\infty \frac{\frac{1}{2\sqrt{t}} dt}{1+(\sqrt{t})^4}$

Simplify further: $= \int_0^\infty \frac{dt}{2t^{1/2}(1+t^2)}.$

Step 2: Simplify Further

Rewriting: $\int_0^\infty \frac{dt}{2t^{1/2}(1+t^2)} = \frac{1}{2} \int_0^\infty \frac{dt}{t^{1/2}(1+t^2)}.$

Let $t^{1/2} = u$, so $t = u^2$ and $dt = 2u du$. When $t = 0$, $u = 0$, and when $t \to \infty$, $u \to \infty$.

Substituting, we get: $\frac{1}{2} \int_0^\infty \frac{dt}{t^{1/2}(1+t^2)} = \frac{1}{2} \int_0^\infty \frac{2u \, du}{u (1+u^4)}.$

Simplify: $= \int_0^\infty \frac{du}{1+u^4}.$

Step 3: Solving the Reduced Integral

Now we compute: $\int_0^\infty \frac{du}{1+u^4}.$

The integral is a standard result from calculus and is solved using techniques involving complex analysis. The result is: $\int_0^\infty \frac{dx}{1+x^4} = \frac{\pi}{2\sqrt{2}}.$

Final Answer:

$\int_0^\infty \frac{dx}{1+x^4} = \frac{\pi}{2\sqrt{2}}.$

Would you like a detailed explanation of the complex analysis step or have any questions?

Related Questions:

1. How is substitution used to simplify integrals involving $x^4$?
2. What are the standard integral results for rational functions with polynomial denominators?
3. Can the integral be evaluated using numerical methods?
4. How does complex analysis come into play for integrals with infinite limits?
5. What other integrals have solutions involving $\pi$ and special functions?

Tip:

When solving integrals with polynomial denominators, look for symmetry, substitution, and potential connections to standard results!