The value of `int_(0)^(oo)(dx)/(1+x^(4))` is equal to

Options

A. `(pi)/(2sqrt2)`
B. `(pi)/(2)`
C. `(pi)/(sqrt2)`
D. `2pisqrt2`

Answer

To solve the integral

\[ I = \int_{0}^{\infty} \frac{dx}{1+x^4} \]

we can use a technique involving symmetry and substitution. Here’s a step-by-step breakdown of the solution:

### Step 1: Split the Integral
We can express the integral in a form that is easier to handle. We can rewrite the integrand as follows:

\[ I = \int_{0}^{\infty} \frac{dx}{1+x^4} = \int_{0}^{\infty} \frac{1}{1+x^4} \, dx \]

### Step 2: Use Substitution
Let’s use the substitution \( x = \frac{1}{t} \). Then, we have:

\[ dx = -\frac{1}{t^2} dt \]

Changing the limits accordingly, when \( x = 0 \), \( t = \infty \) and when \( x = \infty \), \( t = 0 \). Thus, we can rewrite the integral as:

\[ I = \int_{\infty}^{0} \frac{-\frac{1}{t^2}}{1+\left(\frac{1}{t}\right)^4} dt = \int_{0}^{\infty} \frac{1}{t^4 + 1} dt \]

### Step 3: Combine the Integrals
Now we have two expressions for \( I \):

\[ I = \int_{0}^{\infty} \frac{dx}{1+x^4} \quad \text{and} \quad I = \int_{0}^{\infty} \frac{dt}{1+t^4} \]

Adding these two integrals gives:

\[ 2I = \int_{0}^{\infty} \left( \frac{1}{1+x^4} + \frac{1}{1+\left(\frac{1}{x}\right)^4} \right) dx \]

### Step 4: Simplify the Expression
The expression inside the integral simplifies as follows:

\[ \frac{1}{1+x^4} + \frac{1}{1+\frac{1}{x^4}} = \frac{1}{1+x^4} + \frac{x^4}{x^4+1} = \frac{1+x^4}{1+x^4} = 1 \]

Thus, we have:

\[ 2I = \int_{0}^{\infty} 1 \, dx \]

### Step 5: Evaluate the Integral
However, we need to be careful with the limits. The integral diverges, but we can evaluate the original integral using a different method.

### Step 6: Use Residue Theorem
To properly evaluate \( I \), we can use the residue theorem from complex analysis. The poles of the integrand \( \frac{1}{1+x^4} \) are at the fourth roots of unity:

\[ x^4 = -1 \implies x = e^{i\frac{\pi}{4}}, e^{i\frac{3\pi}{4}}, e^{i\frac{5\pi}{4}}, e^{i\frac{7\pi}{4}} \]

The relevant poles in the upper half-plane are \( e^{i\frac{\pi}{4}} \) and \( e^{i\frac{3\pi}{4}} \).

### Step 7: Calculate the Residues
Calculating the residues at these poles and applying the residue theorem will yield:

\[ I = \frac{\pi}{2\sqrt{2}} \]

### Final Result
Thus, the value of the integral is:

\[ I = \frac{\pi}{2\sqrt{2}} \]