Does the following integral diverge or converge: $$\int_1^2{\frac{dx}{\sqrt{16-x^4}}}$$ I tried substituting $x^2$ for $t$ but that didn't seem to make it easier.
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$$\int_1^2\frac{dx}{\sqrt{16-x^4}}=\int_1^2\frac{dx}{\sqrt{4+x^2}\sqrt{2+x}\sqrt{2-x}}$$
It's a square-root divergence and the integral is therefore convergent.
In fact, you can let $u=2-x$, then the integral becomes $$\int_0^1\frac{du}{\sqrt{4+(2-u)^2}\sqrt{4-u}\sqrt{u}}$$
Further let $v=u^2$, the integral becomes $$2\int_0^1\frac{dv}{\sqrt{4+(2-\sqrt{v})^2}\sqrt{4-\sqrt{v}}}$$ which is continuous and bounded in $[0,1]$ and so is clearly integrable.
velut luna
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Could you please elaborate a bit more? Is every integral with a discontinuity and square roots in the denominator convergent? I've seen many of similar exercises but don't have any good general way of figuring it out so if there is a general way I would really appreciate it – Mattias Johnson Oct 25 '17 at 20:17
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Try to integrate $\int \frac{1}{\sqrt{2-x}}dx$ on said interval and see what happens – imranfat Oct 25 '17 at 20:20
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$$\int_{1}^{2}\frac{1}{\sqrt{16-x^4}}\,dx\stackrel{x^4\mapsto z}{=}\frac{1}{4}\int_{1}^{16}\frac{dz}{z^{3/4}\sqrt{16-z}}\stackrel{z\mapsto 16-u^2}{=}\frac{1}{2}\int_{0}^{\sqrt{15}}\frac{du}{(16-u^2)^{3/4}}$$
is converging since it equals the integral of a continuous function over $[0,\sqrt{15}]$.
The exact value of the LHS is given by an incomplete Beta function, namely
$$ \tfrac{1}{8\sqrt{2\pi}}\,\Gamma\left(\tfrac{1}{4}\right)^2-\tfrac{1}{4}\cdot {}_2 F_1\left(\tfrac{1}{4},\tfrac{1}{2};\tfrac{5}{4};\tfrac{1}{16}\right)\approx 0.4039.$$
Jack D'Aurizio
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