Integrate ∫ 0 ∞ x x 2 + 1 d x

Question

Integrate       ∫    0    ∞              x                      x                  2                    +      1              d    x

Lena Gomez · Accepted Answer

We may now use a semicircular contour C in the half plane

ℜ z \geq 0

rather than the annulus as we have disposed of the branch point at the origin. We write

\oint_{C} d z \frac{z^{2}}{z^{4} + 1} = i 2 π ({R e s}_{z = e^{i \frac{π}{4}}} \frac{z^{2}}{z^{4} + 1} + {R e s}_{z = e^{i \frac{3 π}{4}}} \frac{z^{2}}{z^{4} + 1})

\begin{aligned} {R e s}_{z = e^{i \frac{π}{4}}} \frac{z^{2}}{z^{4} + 1} & = \frac{e^{i \frac{2 π}{4}}}{(e^{i \frac{π}{4}} - e^{i \frac{3 π}{4}}) (e^{i \frac{π}{4}} - e^{i \frac{- 3 π}{4}}) (e^{i \frac{π}{4}} - e^{i \frac{- π}{4}})} \\ = \frac{i}{i (1 - i) (2) (2 i \sin \frac{π}{4})} \\ = - i \frac{1 + i}{4 \sqrt{2}} \end{aligned}

Similarly,

{R e s}_{z = e^{i \frac{3 π}{4}}} = - i \frac{1 - i}{4 \sqrt{2}}

Therefore,

\oint_{C} d z \frac{z^{2}}{z^{4} + 1} = i 2 π (- i) \frac{1}{2 \sqrt{2}} = \frac{π}{\sqrt{2}}

Now, about the contour C which has a large radius of, say, R

\oint_{C} d z \frac{z^{2}}{z^{4} + 1} = \int_{C_{R}} d z \frac{z^{2}}{z^{4} + 1} + \int_{- R}^{R} d t \frac{t^{2}}{t^{4} + 1}

In the limit as

R \to \infty

\int_{C_{R}} d z \frac{z^{2}}{z^{4} + 1} \sim \frac{π}{R}

and therefore vanishes. We may then conclude that

\int_{- \infty}^{\infty} d t \frac{t^{2}}{t^{4} + 1} = \int_{0}^{\infty} \frac{\sqrt{x}}{x^{2} + 1} = \frac{π}{\sqrt{2}}

Jared Lowe · Accepted Answer

Though this is not a &quot;complex analysis&quot; solution I can&#039;t resist to post it. We have the following identities  I  =      ∫    0    ∞              x                      x        2            +      1        d  x  =  {  x  →      t    2    }  =  2      ∫    0    ∞              t      2                      t        4            +      1        d  t  =  {  t  →      t          −      1        }  =  2      ∫    0    ∞        1                  t        4            +      1        d  tHence  I  =      ∫    0    ∞                      t        2            +      1                      t        4            +      1        d  t  =      ∫    0    ∞              d      (      t      −              t                  −          1                    )              (      t      −              t                  −          1                            )        2            +      2        d  t  =      ∫          −      ∞        ∞              d      u                      u        2            +      2        =      1          2        arctan  ⁡      (          u              2              )                      |                    −      ∞              ∞        =      π          2

Integrate int_0^infty sqrt(x)/(x^2+1)dx

Answered question

Answer & Explanation

New Questions in Calculus 1