calculate-0-1-tdt-1-t-4-

Question Number 40126 by maxmathsup by imad last updated on 16/Jul/18

c a l c u l a t e \int_{0}^{1} \frac{t d t}{1 + t^{4}}

Commented by maxmathsup by imad last updated on 21/Jul/18

w e h a v e \int_{0}^{\infty} \frac{t}{1 + t^{4}} d t = \int_{0}^{1} \frac{t}{1 + t^{4}} d t + \int_{1}^{+ \infty} \frac{t}{1 + t^{4}} d t b u t

\int_{1}^{+ \infty} \frac{t}{1 + t^{4}} d t =_{t = \frac{1}{x}} \int_{0}^{1} \frac{1}{x (1 + \frac{1}{x^{4}})} \frac{d x}{x^{2}} = \int_{0}^{1} \frac{d x}{x^{3} (1 + \frac{1}{x^{4}})}

= \int_{0}^{1} \frac{d x}{x^{3} + \frac{1}{x}} d x = \int_{0}^{1} \frac{x d x}{x^{4} + 1} \Rightarrow

\int_{0}^{\infty} \frac{t}{1 + t^{4}} d t = 2 \int_{0}^{1} \frac{t d t}{1 + t^{4}} \Rightarrow \int_{0}^{1} \frac{t d t}{1 + t^{4}} = \frac{1}{2} \int_{0}^{\infty} \frac{t}{1 + t^{4}} d t c h a n g e m e n t

t = u^{\frac{1}{4}} g i v e \int_{0}^{\infty} \frac{t}{1 + t^{4}} d t = \int_{0}^{\infty} \frac{u^{\frac{1}{4}}}{1 + u} \frac{1}{4} u^{\frac{1}{4} - 1} d u

= \frac{1}{4} \int_{0}^{\infty} \frac{u^{\frac{1}{2} - 1}}{1 + u} d u = \frac{1}{4} \frac{π}{s i n (\frac{π}{2})} = \frac{π}{4} \Rightarrow

\int_{0}^{1} \frac{t d t}{1 + t^{4}} = \frac{π}{8} .