Integrals

5.7 Integrals

5.7.1 Integrals from 0 to inﬁnity
5.7.2 Integrals from -inﬁnity to inﬁnity

5.7.1 Integrals from 0 to inﬁnity

$\begin{aligned} \int_{0}^{\infty} x^{n} e^{- x} d x & = n! \\ \int_{0}^{\infty} x^{n} e^{- a x} d x & = n! \frac{1}{a^{n + 1}} use y = a x \\ \int_{0}^{\infty} x^{3} e^{- x} d x & = 3! \\ \int_{0}^{\infty} \frac{x^{3}}{e^{x} - 1} d x & = 3! ξ (4) \end{aligned}$

Start by multiplying numerator and denominator by $e^{- x}$ using $\frac{1}{1 - y} = 1 + y + y^{2} + \dots$ which becomes $\int_{0}^{\infty} x^{3} \sum_{n = 1}^{\infty} e^{- n x} d x$ or $\sum_{n = 1}^{\infty} \int_{0}^{\infty} x^{3} e^{- n x} d x$ , then use $z = n x$ , this gives $\sum_{n = 1}^{\infty} \frac{1}{n^{4}} \int_{0}^{\infty} z^{3} e^{- z} d x$ or $(3!) \sum_{n = 1}^{\infty} \frac{1}{n^{4}}$ or $3! ξ (4)$ $\int_{0}^{\infty} e^{- x^{4}} d x = \frac{1}{4} Γ (\frac{1}{4})$ Start by using $x^{4} = y$ or $x = y^{\frac{1}{4}}$ . then $\frac{d y}{d x} = \frac{1}{4} y^{(\frac{1}{4} - 1)}$ , now the integral becomes $\frac{1}{4} \int_{0}^{\infty} y^{(\frac{1}{4} - 1)} e^{- y} d y$ and compare this to $\int_{0}^{\infty} y^{(s - 1)} e^{- x} d x = Γ (s)$ $\int_{0}^{\infty} e^{- \sqrt{x}} d x = \int_{0}^{\infty} e^{- x^{\frac{1}{2}}} d x$ Use same method as above. Will get $2 Γ (2) = 2$ $\begin{aligned} ζ (s) Γ (s) & = \int_{0}^{\infty} \frac{x^{(s - 1)}}{e^{x} - 1} d x s > 1 \\ ζ (n + 1) (n!) & = \int_{0}^{\infty} \frac{x^{n}}{e^{x} - 1} d x n > 0 \\ ζ (2) & = \frac{π^{2}}{6} \\ ζ (4) & = \frac{π^{4}}{90} \\ ζ (s) & = \sum_{n = 1}^{\infty} \frac{1}{n^{s}} s > 1 \\ ζ (4) & = \frac{1}{1^{4}} + \frac{1}{2^{4}} + \frac{1}{3^{4}} + \dots \end{aligned}$

So given $\int_{0}^{\infty} \frac{x^{3}}{e^{x} - 1} d x$ , write as $\int_{0}^{\infty} \frac{x^{(4 - 1)}}{e^{x} - 1} d x = ζ (4) Γ (4)$ or $(3!) ζ (4)$ $\begin{aligned} \int_{0}^{\infty} x^{n} e^{- x} d x & = n! \\ \int_{0}^{\infty} x^{1 - n} e^{- x} d x & = Γ (n) = (n - 1)! \end{aligned}$

$\begin{aligned} I & = \int \frac{d x}{\sqrt{a^{2} - x^{2}}} use x = a \sin θ \\ I & = \int \frac{d x}{x^{2} + a^{2}} use x = a \tan θ \\ I & = \int_{0}^{\infty} x e^{- a x^{2}} d x use u = x^{2} \\ I & = \int_{0}^{\infty} e^{- a x^{2}} d x use I = \frac{1}{2} \int_{- \infty}^{\infty} e^{- a x^{2}} d x = \frac{1}{2} \sqrt{\frac{π}{a}} \end{aligned}$

For $I = \int_{0}^{\infty} x^{n} e^{- a x^{2}} d x$ or $I = \int_{- \infty}^{\infty} x^{n} e^{- a x^{2}} d x$ . If $n$ is even, use the trick of $I (a) = \int_{- \infty}^{\infty} e^{- a x^{2}} d x$ and repeated $I^{'} (a)$ . if $n$ is odd, use $I (a) = \int_{- \infty}^{\infty} x e^{- a x^{2}} d x = \frac{1}{2 a}$ (integration by parts) and then repeated $I^{'} (a)$ .

GAMMA:

$\begin{aligned} Γ (n) & = \int_{0}^{\infty} x^{n - 1} e^{- x} d x \\ Γ (\frac{1}{2}) & = \int_{0}^{\infty} x^{\frac{- 1}{2}} e^{- x} d x \end{aligned}$

use $u = x^{\frac{1}{2}}$ , then $\frac{d u}{d x} = \frac{1}{2} x^{- \frac{1}{2}}$ and the integral becomes $\int_{0}^{\infty} x^{\frac{- 1}{2}} e^{- x} d x = \int_{0}^{\infty} \frac{1}{u} e^{- u^{2}} (2 u d u) = 2 \int_{0}^{\infty} e^{- u^{2}} d u = \sqrt{π}$ $\begin{aligned} I & = \int_{0}^{\infty} x e^{- a x} \sin k x d x \\ I & = \int_{0}^{\infty} x e^{- a x} \cos k x d x \end{aligned}$

For these, we will be given $I = \int_{0}^{\infty} e^{- a x} \sin k x d x$ and then use $I (a) = \int_{0}^{\infty} e^{- a x} \sin k x d x$ and then do the $I^{'} (a)$ method.

5.7.2 Integrals from -inﬁnity to inﬁnity

$\begin{aligned} \int_{- \infty}^{\infty} e^{- x^{2}} d x & = \sqrt{π} \\ \int_{- \infty}^{\infty} e^{- a x^{2}} d x & = \sqrt{\frac{π}{a}} a > 0 \\ \int_{- \infty}^{\infty} e^{- a {(x + b)}^{2}} d x & = \sqrt{\frac{π}{a}} a > 0 \\ \int_{- \infty}^{\infty} x^{n} e^{- a x^{2}} d x & = I for n even, use the I^{'} (a) method \end{aligned}$

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