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Integration by parts. Need to show that:

If $I_n(x)=\int_{0} ^{x}t^n(t^2+a^2)^{-\frac{1}{2}}dt$

Then: $nI_n(x) = x^{n-1}\sqrt{x^2+a^2}-(n-1)a^2I_{n-2(x)}$ if $x\geq2$

I can get to the point where:

$I_n=\frac{x^{n+1}}{n+1}(x^2+a^2)^{-\frac{1}{2}}+\frac{1}{n+1}\int_0 ^x (t^nt^2+a^2t^n-a^2t^n)(t^2+a^2)^{-\frac{3}{2}}dt$

$=~\frac{x^{n+1}}{n+1}(x^2+a^2)^{-\frac{1}{2}}+ \frac{1}{n+1}I_n-\frac{1}{n+1}\int_0 ^x a^2t^n(t^2+a^2)^{-\frac{3}{2}}dt$

$=>~nI_{n}=x^{n+1}(x^2+a^2)^{-\frac{1}{2}}-a^2I_{n-2}+a^4\int t^{n-2}(t^2+a^2)^{-\frac{3}{2}}dt$

Now, is this a good start or should I have taken another route? Because I can not find a way out :/

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1 Answer 1

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Let $$I_n(x) = \int_0^x \dfrac{t^n}{\sqrt{t^2 + a^2}} dt = \int_0^x \dfrac{t^{n-1} \cdot t}{\sqrt{t^2 + a^2}} dt$$ Let $u(t) = t^{n-1}$ and $dv(t) = \dfrac{t}{\sqrt{t^2 + a^2}} dt \implies v(t) = \sqrt{t^2+a^2}$. Hence, \begin{align} I_n(x) & = \int_0^x u(t) dv(t) = \left. u(t) v(t) \right\vert_{0}^x - \int_0^x v(t)du(t)\\ & = x^{n-1} \sqrt{x^2 + a^2} - (n-1)\int_0^x t^{n-2} \sqrt{t^2+a^2} dt \end{align} \begin{align} \int_0^x t^{n-2} \sqrt{t^2+a^2} dt & = \int_0^x \dfrac{t^{n-2} (t^2+a^2)}{\sqrt{t^2+a^2}} dt = \int_0^x \dfrac{t^{n}}{\sqrt{t^2+a^2}} dt + a^2 \int_0^x \dfrac{t^{n-2}}{\sqrt{t^2+a^2}} dt\\ & = I_n(x) + a^2 I_{n-2}(x) \end{align} Plugging this in, we get that $$I_n(x) = x^{n-1} \sqrt{x^2 + a^2} - (n-1)(I_n(x) + a^2 I_{n-2}(x))$$ Rearranging, we get that $$nI_n(x) = x^{n-1} \sqrt{x^2 + a^2} - (n-1)a^2 I_{n-2}(x)$$

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thank you! but it would have been enough just to provide a hint :) –  Sarunas Nov 19 '12 at 7:58

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