Difficult improper integral: $\int_0^\infty \frac{x^{23}}{(5x^2+7^2)^{17}}\,\mathrm{d}x$

How can I find a closed-form expression for the following improper integral in a slick way?

$$\mathcal{I}= \int_0^\infty \frac{x^{23}}{(5x^2+7^2)^{17}}\,\mathrm{d}x$$

• letting $u=x^2$ gives you something slightly simpler, then integrate by partial fractions. – Thomas Andrews Apr 12 '15 at 12:55
• @ThomasAndrews I'm actually looking for a slicker way to do this: integrating by that means a function of such high degree would be very annoying. – math-fun Apr 12 '15 at 12:56
• @math-fun interesting id :-) – Math-fun Apr 12 '15 at 13:00
• @math-fun If you want something specific in an answer, put it in the question. – Thomas Andrews Apr 12 '15 at 13:01
• @ThomasAndrews I've edited it. Thank you. – math-fun Apr 12 '15 at 13:02

Starting from Thomas Andrews's reformulation, repeated integration by parts gives

$$\int_a^\infty{(x-a)^{11}\over x^{17}}dx={11\over16}\int_a^\infty{(x-a)^{10}\over x^{16}}dx=\cdots={11\cdot10\cdots1\over16\cdot15\cdots6}\int_a^\infty{dx\over x^6}={1\over{16\choose5}}{1\over5a^5}$$

so your answer, if I've done all the arithmetic correctly, is

$${1\over2\cdot5^{12}}{1\over{16\choose5}}{1\over5\cdot7^{10}}={1\over2^5\cdot3\cdot5^{13}\cdot7^{11}\cdot13}$$

• Could you be more explicit (at least in the first and the last step) about how you're integrating by parts? (I like the solution, by the way). – math-fun Apr 12 '15 at 13:29
• Ah, nice, that's definitely an easier conclusion to the answer than mine. – Thomas Andrews Apr 12 '15 at 13:29
• $u=(x-a)^{11}$ and $dv=\frac{dx}{x^{17}}$ @math-fun – Thomas Andrews Apr 12 '15 at 13:30
• @math-fun, as Thomas just indicated, use the numerator as the $u$ and the denominator as the $dv$. This decreases the exponent in each by $1$ each time, until the numerator is gone entirely. – Barry Cipra Apr 12 '15 at 13:32
• For what it's worth, this yields the same answer as my messy sum. – Thomas Andrews Apr 12 '15 at 13:42

Let $u=5x^2+49$ then this integral is:

$$\frac{1}{10}\int_{49}^{\infty}\frac{ \left(\frac{u-49}{5}\right)^{11}}{u^{17}}\,du=\frac{1}{2\cdot 5^{12}}\int_{49}^\infty \frac{(u-49)^{11}}{u^{17}}\,du$$

That's going to be messy, but it isn't hard. We get:

$$\frac{(u-49)^{11}}{u^{17}}=\sum_{i=0}^{11}\binom{11}{i}(-49)^{i}u^{-6-i}$$ So an indefinite integral is:

$$\sum_{i=0}^{11} \frac{-1}{5+i}\binom{11}{i}(-49)^i u^{-5-i}$$

which is zero at $\infty$ so we only subtract that case $u=49$ which gives:

$$\int_{49}^\infty \frac{(u-49)^{11}}{u^{17}}\,du = \frac{1}{49^5}\sum_{i=0}^{11}\frac{(-1)^i}{5+i}\binom{11}{i}$$

• Could you please complete the calculation? – math-fun Apr 12 '15 at 13:08
• non realistic method! – RE60K Apr 12 '15 at 13:14
• @math-fun. You have the method(s), you have the result. What else could we do for you ? – Claude Leibovici Apr 12 '15 at 13:14
• @math-fun It's just the binomial expansion of $(x-49)^{11}$ divided by $x^{17}$. – Thomas Andrews Apr 12 '15 at 13:23
• see mine which is more easier – RE60K Apr 12 '15 at 14:16

This is in the form $$J=\int_0^{\infty} \frac{x^{s-1}}{(a+bx^n)^m} \, dx, \tag{1}$$ with $a=49,b=5,n=2,m=17,s=24$. It is clear that this converges To do (1), first change variables to $y=(b/a) x^n$, so $dy/y=n dx/x$, and $$J = \frac{1}{n} \int_0^{\infty} \frac{(a/b)^{s/n}y^{s/n-1}}{a^m(1+y)^m} \, dy = \frac{1}{n}\frac{a^{s/n-m}}{b^{s/n}} \int_0^{\infty} \frac{y^{s/n-1}}{(1+y)^m} \, dy.$$ Now, at this point you can stick the numbers in, and do $\int_0^{\infty} \frac{y^{11}}{(1+y)^{17}} \, dy$, but I'm going to do (1) in general, which is now a matter of evaluating $$J' = \int_0^{\infty} \frac{y^{s/n-1}}{(1+y)^m} \, dy$$

There are still a number of ways to do this: contour integration is a possibility, although problematic if $s/n$ is an integer, as in this case. The easier way turns out to be to write $$\frac{1}{(1+y)^m} = \frac{1}{(m-1)!}\int_0^{\infty} \alpha^{m-1} e^{-(1+y)\alpha} \, d\alpha. \tag{2}$$ (Of course, this generalises to non-integers by using the Gamma function). Putting this into $J'$ and changing the order of integration gives $$J' = \frac{1}{(m-1)!} \int_0^{\infty} \alpha^{m-1}e^{-\alpha} \left( \int_0^{\infty} y^{s/n-1} e^{-\alpha y} \, dy \right) \, d\alpha.$$ Doing the inner integral is just a matter of using (2) again: it is $$\int_0^{\infty} y^{s/n-1} e^{-\alpha y} \, dy = \alpha^{-s/n} (s/n-1)!.$$ Then we just have to do the outer integral, which is $$J' = \frac{(s/n-1)!}{(m-1)!} \int_0^{\infty} \alpha^{m-s/n-1}e^{-\alpha} \, d\alpha = \frac{(s/n-1)!}{(m-1)!}(m-s/n-1)!,$$ applying (2) yet again. Hence the original integral evaluates to $$J = \frac{a^{s/n-m}}{b^{s/n}}\frac{(s/n-1)!(m-s/n-1)}{n(m-1)!} = \frac{a^{s/n-m}}{b^{s/n}} \frac{1}{s} \binom{m-1}{s/n}^{-1} .$$

Sticking the numbers in gives $$\mathcal{I} = \frac{49^{-5}}{5^{12}}\frac{11!4!}{2(16!)},$$ which is easy enough to calculate.

• Just curious ... what inspired you to take this tact rather than the, perhaps, more obvious and straightforward "brute force" approaches? – Mark Viola Apr 13 '15 at 2:58
• @Dr.MV A number of reasons, really: 1. Some generality is a good thing, since it means you only have to do a problem once to get a whole class of solutions, 2. This is a particularly useful integral: I used it myself several times today, 3. The proof is a particular favourite of mine, and 4. I feel more people should know about this trick, since it can often be used to avoid contour integration altogether. And perhaps 5., a bit of showing off. – Chappers Apr 13 '15 at 12:49

$$\mathcal{I}= \int_0^{\infty} \frac{x^{23}}{(5x^2+49)^{17}}\,\mathrm{d}x$$ Let $y=\left(\frac{49b}{1-b}\right)^{1/2}$ which is the sum total of some 4 or 5 substitutions in one go, it is not advisable to use this directly but go with $y=x^2$ then $z=5+49/y$ then $a=1-5/z$ and then $b=1-a$. Note that result could be evaluated at a but the $(1-a)^{11}$ is tedious to expand so I changed the form.Anyways by beta function there is no difference.
Now: $$\mathcal{I}= \frac{1}{2*49^5*5^{12}}\int_{0}^{1} b^{11}(1-b)^4\,\mathrm{d}b$$ Now using beta function or expanding and integrating and adding: $$\int_{0}^{1} b^{11}(1-b)^4\,\mathrm{d}b=\frac{11!*4!}{16!}$$ So answer is: $$\boxed{\displaystyle\large\qquad\mathcal{I}=\frac{11!*4!}{16!*2*49^5*5^{12}}=\frac1{3012333710039062500000}\qquad}$$