Question #9ae3f

1 Answer
Nov 11, 2017

${\int}_{0}^{\infty} \frac{\mathrm{dt}}{{t}^{4} + 1} = \frac{\pi \sqrt{2}}{4}$

Explanation:

$I$=${\int}_{0}^{\infty} \frac{\mathrm{dt}}{{t}^{4} + 1}$

After using $t = \frac{1}{y}$ and $\mathrm{dt} = - \frac{\mathrm{dy}}{y} ^ 2$ transformation,

$I$=${\int}_{\infty}^{0} \frac{- \frac{\mathrm{dy}}{y} ^ 2}{{\left(\frac{1}{y}\right)}^{4} + 1}$

=${\int}_{\infty}^{0} \frac{- {y}^{2} \cdot \mathrm{dy}}{{y}^{4} + 1}$

=${\int}_{0}^{\infty} \frac{{y}^{2} \cdot \mathrm{dy}}{{y}^{4} + 1}$

=${\int}_{0}^{\infty} \frac{{t}^{2} \cdot \mathrm{dt}}{{t}^{4} + 1}$

After collecting 2 integrals,

$2 I$=${\int}_{0}^{\infty} \frac{\left({t}^{2} + 1\right) \cdot \mathrm{dt}}{{t}^{4} + 1}$

=${\int}_{0}^{\infty} \frac{\left({t}^{2} + 1\right) \cdot \mathrm{dt}}{\left({t}^{2} + \sqrt{2} \cdot t + 1\right) \left({t}^{2} - \sqrt{2} \cdot t + 1\right)}$

=$\frac{1}{2} \cdot {\int}_{0}^{\infty} \frac{\left(2 {t}^{2} + 2\right) \cdot \mathrm{dt}}{\left({t}^{2} + \sqrt{2} \cdot t + 1\right) \left({t}^{2} - \sqrt{2} \cdot t + 1\right)}$

=$\frac{1}{2} \cdot {\int}_{0}^{\infty} \frac{\mathrm{dt}}{{t}^{2} + \sqrt{2} \cdot t + 1}$+$\frac{1}{2} \cdot {\int}_{0}^{\infty} \frac{\mathrm{dt}}{{t}^{2} - \sqrt{2} \cdot t + 1}$

=$\frac{1}{2} \cdot {\int}_{0}^{\infty} \frac{2 \cdot \mathrm{dt}}{2 {t}^{2} + 2 \sqrt{2} \cdot t + 2}$+$\frac{1}{2} \cdot {\int}_{0}^{\infty} \frac{2 \cdot \mathrm{dt}}{2 {t}^{2} - 2 \sqrt{2} \cdot t + 2}$

=$\frac{\sqrt{2}}{2} \cdot {\int}_{0}^{\infty} \frac{\sqrt{2} \cdot \mathrm{dt}}{{\left(\sqrt{2} \cdot t + 1\right)}^{2} + 1}$+$\frac{\sqrt{2}}{2} \cdot {\int}_{0}^{\infty} \frac{\sqrt{2} \cdot \mathrm{dt}}{{\left(\sqrt{2} \cdot t - 1\right)}^{2} + 1}$

=$\frac{\sqrt{2}}{2} \cdot {\left[A r c \tan \left(\sqrt{2} \cdot t + 1\right)\right]}_{0}^{\infty}$+$\frac{\sqrt{2}}{2} \cdot {\left[A r c \tan \left(\sqrt{2} \cdot t - 1\right)\right]}_{0}^{\infty}$

=$\frac{\pi \sqrt{2}}{2}$

Thus, $I = \frac{\pi \sqrt{2}}{4}$