definite integral #2

**Random Variable** · January 13th 2011, 07:51 PM

$\displaystyle \int^{3}_{0} \frac{\big(x^{3}(3-x)\big)^{\frac{1}{4}}}{5-x} \ dx$

There is a worked solution on Wikipedia using contour integration. I find it quite confusing.

Methods of contour integration - Wikipedia, the free encyclopedia

However, you don't have to resort to contour integration to solve the problem. There's a much easier (albeit longer) way.

**Unbeatable0** · January 14th 2011, 02:36 AM

Is this what you're after?

Let $x=\frac{3}{1+t^4}$ , then the integral equals:

$\displaystyle{<br /> 18\int_{-\infty}^\infty \frac{t^4}{(t^4+1)^2(5t^4+2)} dt<br /> }$

and we know this integral is equal to $2\pi i$ times the sum of residues of the integrand in the upper half of the complex plane, which is simple, technical and tedious to calculate.

Alternatively, this can be calculated with partial fractions, with the help of the following identity:

$\displaystyle{<br /> \int_0^\infty \frac{1}{1+x^a}dx = \frac{\pi}{a}\csc\frac{\pi}{a}<br /> }$

**Random Variable** · January 14th 2011, 08:45 AM

Is this what you're after?

Not exactly.

Rewrite the integral as $\displaystyle \int^{3}_{0} \frac{x}{5-x} \Big(\frac{3-x}{x}\Big)^{\frac{1}{4}}$

let $t = \frac{3-x}{x}$

then the integral becomes $\displaystyle 9 \int_{0}^{\infty} \frac{1}{(5t+2)(t+1)^{2}} \ t^{\frac{1}{4}} \ dt$

use partial fractions

$\displaystyle 25 \int^{\infty}_{0} \frac{t^{\frac{1}{4}}}{5t+2} \ dt t - 5 \int^{\infty}_{0} \frac{t^{\frac{1}{4}}}{t+1} \ dt - 3 \int^{\infty}_{0} \frac{t^{\frac{1}{4}}}{(t+1)^{2}} \ dt$

Now write each integral as a double integral.

$= \displaystyle 25 \int^{\infty}_{0} \int^{\infty}_{0} t^{\frac{1}{4}} \ e^{-(5t+2)y} \ dy \ dt - 5 \int^{\infty}_{0} \int^{\infty}_{0} t^{\frac{1}{4}} \ e^{-(t+1)y} \ dy \ dt- 3 \int^{\infty}_{0} \int^{\infty}_{0} \ t^{\frac{1}{4}} \ y \ e^{-(t+1)y} \ dy \ dt$

From here swtich the order of integration, integrate, and then use Euler's reflection formula and the property $\Gamma(z+1) = z \Gamma(z)$ .

After some final manipulation/simplification, you'll get the answer.

**Random Variable** · January 14th 2011, 09:17 AM

Wait a minute. The first two integrals after using partial fractions diverge. Is it even legal to do what I did next? How did I I get the correct answer?

**SammyS** · January 14th 2011, 12:03 PM

Originally Posted by Random Variable

Wait a minute. The first two integrals after using partial fractions diverge. Is it even legal to do what I did next? How did I I get the correct answer?

Yes, but summing the integrands of those two gives an integral which does converge.

$\displaystyle 5 \int^{\infty}_{0} t^{\frac{1}{4}}\left( \frac{5}{5t+2} - \frac{1}{t+1}\right) \ dt$

$\displaystyle =5 \int^{\infty}_{0} \frac{3t^{\frac{1}{4}}}{(5t+2)(t+1)} \ dt$

**Random Variable** · January 14th 2011, 12:23 PM

It's valid to switch the order of integration of the first two double integrals since the integrands are always positive. The issue is what exactly is $\Gamma\Big(\frac{5}{4}\Big)\int^{\infty}_{0} y^{-5/4}e^{-y} \ dy$ . Is it infinity because the integral diverges? Or is it $\Gamma\Big(\frac{5}{4}\Big) \Gamma\Big(\frac{-1}{4} \Big) = \Gamma\Big(\frac{5}{4} \Big)\Gamma\Big(1-\frac{5}{4} \Big) = \frac{\pi}{\sin \frac{5 \pi}{4}}} = - \sqrt{2} \pi$ ?

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