interesting Integral , alternative solution.

Question

Show the following relation:

$$\int_{0}^{\infty} \frac{x^{29}}{(5x^2+49)^{17}} \,\mathrm dx = \frac{14!}{2\cdot 49^2 \cdot 5^{15 }\cdot 16!}.$$

I came across this intgeral on a physics forum and solved by (1) making a substitution (2) finding a recursive formula (with integration by parts).

I would like to see a clever alternative solution to this one, since the only way I would solve such an integral is by finding a recursive formula. But I'm sure there are other ways/methods to do it (maybe more elegant) that I want to discover.

Answer 1

Not sure what you consider elegant or done by substitution, but the obvious way to attack this integral in my eyes is to first substitute $y=x^2$ and then $u=5 y+49$ to get the integral

$$\frac12 \frac{1}{5^{15}} \int_{49}^{\infty} du \frac{(u-49)^{14}}{u^{17}}$$

This may be evaluated by applying the binomial theorem becomes

$$\frac12 \frac{1}{5^{15}} \sum_{k=0}^{14} (-1)^k \binom{14}{k} 49^{14-k} \int_{49}^{\infty} \frac{du}{u^{17-k}} = \frac12 \frac{1}{5^{15}} \frac{1}{49^2} \sum_{k=0}^{14} (-1)^k \frac{\binom{14}{k}}{16-k}$$

The sum on the right may be evaluated by manipulating the binomial coefficient as follows:

$$\binom{14}{k} \frac{1}{16-k} = \frac{14!}{16!} \binom{16}{k} (15-k)$$

$$\sum_{k=0}^N (-1)^k \binom{N}{k} = 0$$

$$k \binom{N}{k} = N \binom{N-1}{k-1}$$

to find that

$$\sum_{k=0}^{14} (-1)^k \frac{\binom{14}{k}}{16-k} = \frac{14!}{16!}$$

and the desired result is attained.

Answer 2

The given integral is \begin{align} I=&\int_{0}^{\infty}\frac{x^{29}}{(5x^2+49)^{17}}dx\\ \ =&\frac{1}{2\cdot 5^{15}}\int_{0}^{\infty}\frac{z^{14}}{(z+49)^{17}}dz\quad(\mbox{substituting}\ 5x^2=z)\\ \ =&\frac{1}{2\cdot 5^{15}}\int_{0}^{\pi/2}\frac{(49)^{14}\tan^{28}\theta\cdot2\cdot49\cdot\tan \theta\sec^2\theta}{49^{17}\sec^{34}\theta}dz\quad(\mbox{substituting}\ z=49\tan^2\theta)\\ \ =&\frac{1}{2\cdot 5^{15}\cdot49^2}\cdot2\int_{0}^{\pi/2}\frac{\sin^{29}\theta}{\cos^{29}\theta}\cos^{32}\theta\ d\theta\\ \ =&\frac{1}{2\cdot 5^{15}\cdot 49^2}\cdot\beta\left(15,2\right)\\ \ =&\frac{\Gamma(15)\Gamma(2)}{2\cdot5^{15}\cdot49^2\cdot\Gamma(17)}\\ \ =&\frac{14!}{2\cdot5^{15}\cdot49^2\cdot16!} \end{align}

Along similar lines of argument, the integral $I(a,b,c,d)$($a,b,c,d$ positive integres) as defined by Hagen von Eitzen becomes(if I have not committed any mistake) $$\frac{1}{2\cdot c^d}\left(\frac{c}{b}\right)^{(n+1)/2}\beta\left(\frac{n+1}{2},d-\frac{n-1}{2}-1\right)\\ =\frac{1}{2\cdot c^{d-\frac{n+1}{2}}\cdot b^{\frac{n+1}{2}}}\cdot\frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(d-1-\frac{n-1}{2}\right)}{(d-1)!}$$

Answer 3

$$I_{(n,m)}=\int_0^\infty \frac{x^n}{(ax^2+b)^m} \,\mathrm dx$$ $$=x^{n-1}\int_0^\infty\frac{xdx}{(ax^2+b)^m}-\int_0^\infty\left(\frac{d(x^{n-1})}{dx}\int\frac{xdx}{(ax^2+b)^m}\right)dx $$ $$=x^{n-1}\frac1{-2a(m-1)(ax^2+b)^{m-1}}\big|_0^\infty-\frac{n-1}{-2a(m-1)}\int_0^\infty \frac{x^{n-2}}{(ax^2+b)^{m-1}}dx$$

$\lim_{x\to\infty}x^{n-1}\frac1{(ax^2+b)^{m-1}}=\lim_{x\to\infty}\frac1{aO(x^{2m-2-(n-1)})}=0$ if $2m-2-(n-1)=2m-n-1\ge1\iff n\le 2m-2$

$$\implies I_{(n,m)}=\frac{n-1}{2a(m-1)}I_{(n-2,m-1)}\text{if } n\le 2m-2$$

We have $n=29,m=17$ So, we start with $n=2m-5<2m-2$

$$\implies I_{(2m-5,m)}=\frac{2m-5-1}{2a(m-1)}I_{(2m-5-2,m-1)}=\frac{2m-3}{a(m-1)}I_{(2m-7,m-1)}$$

Replacing $m$ with $m-1,$

$$\implies I_{(2m-5,m)}=-\frac{2m-5-1}{2a(m-2)}I_{(2m-5-2,m-1)}=-\frac{m-3}{a(m-2)}I_{(2m-7,m-1)}$$

$$m=17\implies I_{(29,17)}=-\frac{14}{16a}I_{(27,16)}$$

$$m=16\implies I_{(27,16)}=-\frac{13}{15a}I_{(25,15)}$$

$$\cdots$$

$$m=4\implies I_{(3,4)}=-\frac1{3a}I_{(1,3)}$$

Now, $$I_{(1,3)}=\int_0^\infty \frac x{(ax^2+b)^3}dx=\frac1{2a}\int_b^\infty \frac {du}{u^3}=\frac{1}{2a(-2)u^2}\big|_b^\infty=\frac1{4ab^2}$$

$$\implies I_{(29,17)}=\frac{14}{(16\cdot15\cdots4\cdot3)\cdot a^{14}\cdot 4ab^2}=\frac{14!2!}{16!\cdot4a^{15}b^2}=\frac{14!}{16!\cdot2a^{15}b^2}$$

Answer 4

Denote your integral by $J$. By means of the substitution $u:=5x^2+49$, $\>du=10 x\>dx$ we get $$J={1\over 2\cdot 5^{15}}\int_{49}^\infty{(u-49)^{14}\over u^{17}}\ du\ .\tag{1}$$ Consider now for given $a>0$ (we have $a=49$ in mind) the quantities $$Q_n:=(n+2)(n+1)\int_a^\infty{(u-a)^n\over u^{n+3}}\ du\qquad(n\geq0)\ .$$ When $n\geq1$ partial integration gives $$Q_n=-{(n+1)(u-a)^n\over u^{n+2}}\biggr|_a^\infty + (n+1)n\int_a^\infty{(u-a)^{n-1}\over u^{n+2}}\ du=Q_{n-1}\ .$$ It follows that $$Q_n=Q_0=2\int_a^\infty{du\over u^3}={1\over a^2}\ .$$ Plugging this with $n:=14$ and $a:=49$ into $(1)$ we obtain $$J={1\over 2\cdot 5^{15}}\cdot{Q_{14}\over16\cdot 15}={1\over 49^2\cdot5^{15}\cdot480}\ ,$$ as stated in the source.

Answer 5

This solution uses only elementary substitutions, and is based on a post at a greek website by Seraphim. Out of curiosity, where did you find this integral? To get the ball rolling we use the substitution $u = x^2 \ \Rightarrow \ \,\mathrm{d}u = 2x \,\mathrm{d}x $. \begin{align*} I = \int_{0}^{\infty} \frac{x^{29}}{\left( 5x^2 + 49 \right)^{17}} \,\mathrm{d}x = \frac{1}{2}\int_{0}^{\infty} \frac{\left(x^{2}\right)^{14} }{\left( 5x^2 + 49 \right)^{17}} 2x \,\mathrm{d}x = \frac{1}{2}\int_{0}^{\infty} \frac{u^{14} }{\left( 5u + 49 \right)^{17}} \,\mathrm{d}u \end{align*} We can simplify the denominator using $u = \frac{49}{5}t \Rightarrow \,\mathrm{d}u = \frac{49}{5} \,\mathrm{d}t$ which gives \begin{align*} I & = \frac{1}{2}\int_{0}^{\infty} \frac{\left( \frac{49}{5}t\right)^{14} }{\left( 5\left( \frac{49}{5}t\right) + 49 \right)^{17}}\left( \frac{49}{5} \right)\,\mathrm{d}t \\ & = \frac{1}{2} \left( \frac{49}{5}\right)^{14} \left( \frac{49}{5} \right) \int_{0}^{\infty} \frac{ t^{14} }{49^{17} ( t + 1 )^{17}}\,\mathrm{d}t \\ & = \frac{1}{2 \cdot 5^{15} \cdot 49^2} \int_{0}^{\infty} \frac{ t^{14} }{ ( t + 1 )^{17}}\,\mathrm{d}t \\ & = \frac{1}{2 \cdot 5^{15} \cdot 49^2} \int_0^\infty \left( \frac{ t}{ t + 1 } \right)^{14} \frac{\mathrm{d}t}{(t+1)^3} \end{align*} Let $y = \cfrac{t}{t+1}\,, \quad \cfrac{\,\mathrm{d}y}{\,\mathrm{d}t} = \cfrac{1}{(t+1)^2} \Rightarrow \,\mathrm{d}y = \cfrac{1}{(t+1)^2} \,\mathrm{d}t$. This simplifies the integral down to \begin{align*} I & = \frac{1}{2 \cdot 5^{15} \cdot 49^2} \int_{0}^{\infty} \frac{ t^{14} }{ ( t + 1 )^{17}}\,\mathrm{d}t \\ & = \frac{1}{2 \cdot 5^{15} \cdot 49^2} \int_{0}^{\infty} \overbrace{\left( \frac{ t}{ t + 1 } \right)^{14}}^{y} \overbrace{\left( \frac{1}{t+1}\right)}^{1-y}\overbrace{\left(\frac{\mathrm{d}t}{(t+1)^2}\right) }^{\,\mathrm{d}y}\\ & = \frac{1}{2 \cdot 5^{15} \cdot 49^2} \int_{0}^{1} y^{14}(1-y) \,\mathrm{d}y \\ & = \frac{1}{2 \cdot 5^{15} \cdot 49^2} \left[ \frac{1}{15}y^{15} - \frac{1}{16}y^{16} \right]_0^1 \\ & = \frac{1}{2 \cdot 5^{15} \cdot 49^2 \cdot 15 \cdot 16} \\ \int_{0}^{\infty} \frac{x^{29}}{\left( 5x^2 + 49 \right)^{17}} \,\mathrm{d}x& = \frac{14!}{2 \cdot 5^{15} \cdot 49^2 \cdot 16!} \end{align*} Which is what we wanted to show. $\square$

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{29} \over \pars{5x^{2} + 49}^{17}}\,\dd x = {14! \over 2\cdot 49^{2} \cdot 5^{15}\cdot 16!}}:\ {\Large ?}}$.

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\begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{29} \over \pars{5x^{2} + 49}^{17}}\,\dd x} \,\,\,\stackrel{5x^{2}/49\ \mapsto\ x}{=}\,\,\, {1 \over 2\cdot 49^{17}}\,\pars{49 \over 5}^{15} \int_{0}^{\infty}{x^{\color{red}{15} - 1} \over \pars{1 + x}^{17}}\,\dd x \end{align} It can be evaluated by means of the Ramanujan's Master Theorem. Namely, \begin{align} &{1 \over \pars{1 + x}^{17}} = \sum_{k = 0}^{\infty}{-17 \choose k}x^{k} = \sum_{k = 0}^{\infty}{16 + k \choose k}\pars{-1}^{k}x^{k} \\[5mm] = &\ \sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{17 + k} \over 16!}{\pars{-x}^{k} \over k!} \end{align} and \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{29} \over \pars{5x^{2} + 49}^{17}}\,\dd x} = {1 \over 2\cdot 49^{2} \cdot 5^{15}}\, \Gamma\pars{\color{red}{15}}\, {\Gamma\pars{17 - \color{red}{15}} \over 16!} \\[5mm] = &\ {1 \over 2\cdot 49^{2} \cdot 5^{15}}\,14!\,{1! \over 16!} = \bbx{{14! \over 2\cdot 49^{2} \cdot 5^{15}\cdot 16!}} \approx 2.8433 \times 10^{-17} \\ & \end{align}