<s>[INST] Find $\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x$ How can we find the integral:
$$\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x$$
I tried to find and got it to be $\cfrac{\pi}{\sqrt2}$. Am I correct? Please help me with an appropriate method. I tried to use sum of residue.
 [/INST] First, use the change of variable $x=\frac{1}{u}  => dx = -\frac{1}{u^2}du $
$I = \int_{ - \infty }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x = 2*\int_{ 0 }^{ + \infty } {\frac{1} {1 + {x^4}}} \;{\mathrm{d}}x = 2*\int_{ 0 }^{ + \infty } {\frac{u^2} {1 + {u^4}}} \;{\mathrm{d}}u $
=> $2*I = 2*\int_{ 0 }^{ + \infty } {\frac{1+x^2} {1 + {x^4}}} \;{\mathrm{d}}x => I= \int_{ 0 }^{ + \infty } {\frac{1+x^2} {1 + {x^4}}} \;{\mathrm{d}}x $
Now: Let: $t= x-\frac{1}{x}$ => $dt= \frac{1+x^2}{x^2}dx $ x->0, t->$-\infty$ ; x->$+\infty$ , t->$+\infty$
=> $ I = \int_{ - \infty }^{ + \infty } {\frac{1+x^2} {1 + {x^4}}*\frac{x^2}{1+x^2}} \;{\mathrm{d}}t = \int_{ - \infty }^{ + \infty } {\frac{1} {x^{-2} + {x^2}}} \;{\mathrm{d}}t$
$ x^2 + x^{-2} = (x-\frac{1}{x})^2 +2 = t^2 +2 $
=> $ I = \int_{ - \infty }^{ + \infty } {\frac{1} {2 + {t^2}}} \;{\mathrm{d}}t $
Let : t = $\sqrt{2}*v$
$I = \frac{1}{2}*\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {(\frac{t}{\sqrt{2}})^2}}} \;{\mathrm{d}}t = \frac{1}{\sqrt{2}}*\int_{ - \infty }^{ + \infty } {\frac{1} {1 + {v^2}}} \;{\mathrm{d}}v = \frac{\pi}{\sqrt{2}}$
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