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I'm working on a physics problem and I got to the integral:

$$\int_0^\infty (a+b+x^2)^{-\frac{3}2} dx = \frac{1}{(a+b)}$$

I am just trying to understand how this is achieved. Because the indefinite integral yields

$$x*(a+b)^{-1}*(a+b+x^2)^{-\frac{1}2}$$

Evaluating this from 0 to $\infty$, to me, gives

$$\frac{\infty}{\sqrt{a+b+\infty^2}} - 0$$

Edit: corrected my math

abyssmu
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2 Answers2

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Let $t = \frac x{\sqrt{a+b}}$ to reexpress the integral as,

$$I=\int_0^\infty (a+b+x^2)^{-\frac{3}2} dx = \frac{1}{a+b} \int_0^\infty \frac{dt}{(1+t^2)^{3/2}}$$

Then, let $t=\sinh u$

$$I= \frac{1}{a+b} \int_0^\infty \frac{\cosh u}{\cosh^3 u}du =\frac{1}{a+b}\int_0^\infty \text{sech}^2 u\> du=\frac{1}{a+b}\tanh u|_0^\infty=\frac{1}{a+b}$$

Quanto
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The indefinite integral of $(C+x^2)^{-\frac{3}{2}}$ actually equals $\frac{x}{C\sqrt{C+x^2}}$. If we set $C=a+b$, the answer will coincide with yours.

I guess that in your approach you missed $x$ coming from deriving $x^2$ inside brackets.

Aleksandr Logunov
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  • Thank you. I corrected that. My next question regards the infinity/sqrt(c+infinity^2). Is the reason it becomes 1/(a+b) because as you get arbitrarily large, it is dominated by x, and essentially becomes infinity/sqrt(infinity^2) which would be 1?? – abyssmu Dec 10 '19 at 22:44
  • Yes, x dominates here, to find the limit we just need to rewrite the fraction properly: x/(Csqrt(C+x^2)) = (x/x)/(Csqrt(C+x^2)/x) = 1/(C*sqrt(C/x^2+1)) which tends to 1/C for x approaching infinity. – Aleksandr Logunov Dec 10 '19 at 22:52