### Instruction:
Evaluate

$$
I=\int_{0}^{\infty} \frac{d x}{4+x^{4}} .
$$

### Response:

As

$$
y=1+2 x+3 x^{2}+4 x^{3}+\ldots+n x^{n-1}+\ldots
$$

we have

$$
x y=x+2 x^{2}+3 x^{3}+4 x^{4}+\ldots+n x^{n}+\ldots
$$

and thus

$$
y-x y=1+x+x^{2}+x^{3}+\ldots+x^{n-1}+\ldots=\frac{1}{1-x}
$$

since $|x|<1$. Hence

$$
y=\frac{1}{(1-x)^{2}} .
$$

The mean value of $x$ is

$$
\bar{x}=\frac{\int_{0}^{\infty} x f(x) d x}{\int_{0}^{\infty} f(x) d x}=\frac{\int_{0}^{\infty} x^{2} e^{-x / \lambda} d x}{\int_{0}^{\infty} x e^{-x / \lambda} d x}=\frac{2 \lambda \int_{0}^{\infty} x e^{-x / \lambda} d x}{\int_{0}^{\infty} x e^{-x / \lambda} d x}=2 \lambda,
$$

where integration by parts has been used for the numerator and $\lambda>0$ has been assumed . The most probable value of $x, x_{m}$, is given by

$$
\left(\frac{d t}{d x}\right)_{x_{m}}=0,
$$

whence $x_{m}=\lambda$.

Note that if $\lambda<0, f(x)$ is infinite at $\infty$ and the mean and most probable values are also infinite.

Fig. $3.14$Write

$$
\begin{aligned}
I &=\frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{4}+4}+\frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{4}+4} \\
&=\frac{1}{2} \int_{-\infty}^{0} \frac{d x}{x^{4}+4}+\frac{1}{2} \int_{0}^{\infty} \frac{d x}{x^{4}+4}=\frac{1}{2} \int_{\infty}^{\infty} \frac{d x}{x^{4}+4}
\end{aligned}
$$

and consider

$$
\oint_{C} \frac{d z}{z^{4}+4}
$$

where $C$ is the closed contour of Fig. $3.14$ consisting of the line from $-R$ to $R$ and the semicircle $\Gamma$, traversed in anticlockwise sense, in the complex plane. As

$$
\frac{1}{z^{4}+4}=\frac{1}{\left(z-z_{1}\right)\left(z-z_{2}\right)\left(z-z_{3}\right)\left(z-z_{4}\right)},
$$

where

$$
z_{1}=1+i, \quad z_{2}=-1+i, \quad z_{3}=-1-i, \quad z_{4}=1-i
$$

the residue theorem gives

$$
\begin{aligned}
&\oint_{C} \frac{d z}{z^{4}+4} \\
&=2 \pi i\left[\frac{1}{\left(z_{1}-z_{2}\right)\left(z_{1}-z_{3}\right)\left(z_{1}-z_{4}\right)}+\frac{1}{\left(z_{2}-z_{1}\right)\left(z_{2}-z_{3}\right)\left(z_{2}-z_{4}\right)}\right] \\
&=2 \pi i\left[\frac{1}{2(2+2 i) \cdot 2 i}-\frac{1}{2 \cdot 2 i(-2+2 i)}\right]=\frac{\pi}{4} .
\end{aligned}
$$

But

$$
\oint_{C} \frac{d z}{z^{4}+4}=\int_{-R}^{R} \frac{d x}{x^{4}+4}+\int_{\Gamma} \frac{d z}{z^{4}+4} \rightarrow \int_{-\infty}^{\infty} \frac{d x}{x^{4}+4}
$$

as $R \rightarrow \infty$. Hence

$$
I=\frac{1}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8} .
$$