interesting Integral , alternative solution. 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.
 A: 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) $$\frac{1}{2\cdot c^d}\left(\frac{c}{b}\right)^{(n+1)/2}\beta\left(\frac{n+1}{2},d-\frac{n-1}{2}-1\right)\\ =\frac{1}{2\cdot c^{d-\frac{n+1}{2}}\cdot b^{\frac{n+1}{2}}}\cdot\frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(d-1-\frac{n-1}{2}\right)}{(d-1)!}$$
A: 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.
A: $$I_{(n,m)}=\int_0^\infty \frac{x^n}{(ax^2+b)^m} \,\mathrm dx$$
$$=x^{n-1}\int_0^\infty\frac{xdx}{(ax^2+b)^m}-\int_0^\infty\left(\frac{d(x^{n-1})}{dx}\int\frac{xdx}{(ax^2+b)^m}\right)dx $$
$$=x^{n-1}\frac1{-2a(m-1)(ax^2+b)^{m-1}}\big|_0^\infty-\frac{n-1}{-2a(m-1)}\int_0^\infty \frac{x^{n-2}}{(ax^2+b)^{m-1}}dx$$
$\lim_{x\to\infty}x^{n-1}\frac1{(ax^2+b)^{m-1}}=\lim_{x\to\infty}\frac1{aO(x^{2m-2-(n-1)})}=0$ 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, $$I_{(1,3)}=\int_0^\infty \frac x{(ax^2+b)^3}dx=\frac1{2a}\int_b^\infty \frac {du}{u^3}=\frac{1}{2a(-2)u^2}\big|_b^\infty=\frac1{4ab^2}$$
$$\implies I_{(29,17)}=\frac{14}{(16\cdot15\cdots4\cdot3)\cdot a^{14}\cdot 4ab^2}=\frac{14!2!}{16!\cdot4a^{15}b^2}=\frac{14!}{16!\cdot2a^{15}b^2}$$
A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\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 ?}}$.

\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}
