Integrating a function from negative infinity to infinity

Question

i got this weird question about integration from infinity to infinity.

$$\int_{-\infty}^{\infty} \frac{1}{z^2+25}$$

First idea was to take the factor out to get $(z+5)(z-5)$ but that really did not achieve anything.

Then i tried let $x= z^2$ and the $\frac{dx}{dz}= 2z$ then dz = 1/2z but that also went nowhere

Then i tried..... $x=z$, $\frac{dx}{dz}= 1$ hence $dz = dx$

$$\int_{-\infty}^{\infty} \frac{dx}{x^2+25}$$ which has not achieved anything. Please help!

Wait i just remember how this applies to partial fractions. But still confused how to grom infinity to infinity.

So start wit $$\frac{1}{z^2+25}=\frac{1}{(z+5)(z-5)}$$ $$\frac{1}{(z+5)(z-5)}=\frac{A}{x+5}+\frac{B}{x-5}$$ $$1=A(x-5)+B(x+5)$$ At x = 5 $1=10B$ hence $B=0.1

At x=-5 $A=1/-10$

But i'm not sure where to go from there

Answer 1

$$ \int \frac{dz}{z^2+25} = \frac 1 5\int \frac{dz/5}{(z/5)^2+1} = \frac 1 5 \int \frac{du}{u^2+1} = \frac 1 5 \arctan u + C. $$ As $z\to\pm\infty$ then $u\to\pm\infty$, so recall from trigonometry that $\arctan u\to\pm\dfrac\pi2$ as $u\to\pm\infty$.

$$ \frac 1 5 \left( \frac\pi2 - \frac{-\pi}{2} \right) = \frac \pi 5. $$

Answer 2

hint: use the relation $1+\tan^2 x = \tan' x$. Define $z = 5\tan u$.

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Solution: With this change of variables, $dz = 5\tan ' u du$.

$$ \int \frac{dz}{25 + z^2} = \int \frac{5\tan ' u\ du}{25 + 25 \tan^2 u} = \frac u5 \\\implies \int_{\Bbb R} \frac {dz}{25 + z^2} = \left[\frac u5\right]_{-\pi / 2}^{\pi / 2} = \frac {\pi}5 $$

Answer 3

Substituting $$x = 5 \tan \theta$$ transforms the integral to $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{5 \sec^2 \theta \, d\theta}{25 \tan^2 \theta + 25},$$ which simplifies to

$$\displaystyle \frac{1}{5} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta.$$

Rearranging the substitution gives $\theta = \arctan \frac{x}{5}$, and the new limits follow immediately from the limit $$\lim_{u \to \pm \infty} \arctan u = \pm \frac{\pi}{2} .$$

Answer 4

$$\int_{-\infty}^{\infty} \frac{1}{z^2+25}=\int_{-\infty}^{0} \frac{1}{z^2+25}+\int_{0}^{\infty} \frac{1}{z^2+25}=\\ \lim_{x\to -\infty}\int_{x}^{0} \frac{1}{z^2+25}+\lim_{x\to \infty}\int_{0}^{x} \frac{1}{z^2+25}$$ To answer your confusion about infinity.

Answer 5

I'm pretty sure you're supposed to split the bounds. Pick a value between $-\infty$ and $+\infty$ that is valid. In this case $0$ would work (however, that is not always the case). This will give you 2 separate integrals:

$$ \int_{-\infty}^0 \frac{1}{z^{2} + 25} dx \quad\text{and}\quad \int_0^\infty \frac{1}{z^{2} + 25} dx$$

Then just sum up the two integrals, or in terms of a formula:

$$ \int_{-\infty}^\infty \frac{1}{z^2 + 25} dx = \int_{-\infty}^0 \frac{1}{z^{2} + 25} \, dx + \int_0^\infty \frac{1}{z^2 + 25} \, dx$$

As for solving the integral, I don't think you need to apply partial fractions. There should be a nice formula that involves $\tan(\theta)$.

Hope that helps!

EDIT: Joao pointed out a step I missed. You need to transform

$\displaystyle \int_{-\infty}^{0}\,\,$ to $\,\,\displaystyle \lim_{x \to -\infty} \int_{x}^{0}$

Answer 6

I think i finally got it let $z=5tan(x)$ hence $dz=5sec^2(x)dx$ sub this in to the integrand

$$\int_{-\infty}^{\infty} \frac{dz}{z^2+25} =\int_{-\infty}^{\infty} \frac{5sec^2(x)dx}{25tan^2(x)+25}$$

as sec^2(x)=1+tan^2(x)

$$\int_{-\infty}^{\infty} \frac{5(tan^2x+1)}{25(tan^2(x)+1)}= \int_{-\infty}^{\infty} \frac{1dx}{5}$$

$$\int_{-\infty}^{\infty} \frac{dz}{z^2+25} = \frac{x}{5}$$ for z =-$\infty,\infty$

as $$z = 5tan(x)$$ hence $$x=arctan(z/5)$$

observing its graph we find that as x approach positive and negative infinity artctan(z) equal $\frac{\pi}{2},-\frac{\pi}{2}$

this is the same for $x=arctan(z/5)$

hence, $$\int_{-\infty}^{\infty} \frac{dz}{z^2+25} =\frac{1}{5}(\frac{\pi}{2}-(-\frac{\pi}{2}) = \frac{\pi}{5}$$

Answer 7

I'll give an approach different from all of the above using complex analysis, which saves you from a lot of trigonometry and elementary algebraic manipulations. (This might be more likely the preferred approach of the problem, since it uses the variable $z$, which commonly denotes a complex variable.)

Note that the denominator $z^2+25=(z+5i)(z-5i)$ contains complex roots. Hence the integral is susceptible to the technique of integrating over a closed curve on the complex plane and applying the residue theorem.

More specifically, consider the half-circle $\gamma$ on the complex plane with the straight edge sitting on the real interval $[-R,R]$ and with the arc $Re^{i\theta}$ where $\theta\in[0,\pi]$. The function is rational, hence meromorphic, and the root $z=5i$ (with multiplicity $1$) is included inside this half-circle when $R>5$. Hence we compute the residue at $z=5i$:

$$\operatorname{res}_{5i}f=\lim_{z\to5i}\frac{z-5i}{(z+5i)(z-5i)}=\frac{1}{10i}$$

Now apply the residue theorem:

$$\int_\gamma f(z)\,dz=2\pi i\operatorname{res}_{5i}f=\frac{2\pi i}{10i}=\frac{\pi}{5}$$

It remains to prove that the integral around the arc $\gamma_R$ goes to $0$ as $R\to\infty$. The function is bounded on this arc by $1/(R^2+25)$, and the arc has length $\pi R$, so by the estimation lemma:

$$\lim_{R\to\infty}\left|\int_{\gamma_R}f(z)\,dz\right|\le\lim_{R\to\infty}\left|\frac{\pi R}{R^2+25}\right|$$

The right hand side has growth rate $O(R)$ in the numerator and $O(R^2)$ in the denominator, so it's straightforward to show that it goes to $0$ as $R\to\infty$ (by e.g. L'Hôpital's rule). Hence the integral on the real line is equal to the integral around the half-circle, that is

$$\int_{-\infty}^{\infty}\frac{1}{z^2+25}\,dz=\frac{\pi}{5}$$

as was required.