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

  • 4
    $\begingroup$ Note that $z^2+25=(z+5i)(z-5i)\ne (z+5)(z-5)$. One potential substitution could be $z=5\tan\theta$. $\endgroup$
    – abiessu
    Oct 30, 2014 at 1:15
  • 2
    $\begingroup$ Instead of bothering with tedious trigonometric substitutions, one might consider an application of the residue theorem. $\endgroup$
    – Gahawar
    Oct 30, 2014 at 1:36
  • 2
    $\begingroup$ Your integral is missing a $dz$... $\endgroup$
    – David Z
    Oct 30, 2014 at 3:44
  • $\begingroup$ @Gahawar I agree wholeheartedly. Note also that the variable is named $z$, which is usually used to denote a complex variable. I've written up a solution using the residue theorem below. $\endgroup$
    – Marc
    Oct 30, 2014 at 6:06

7 Answers 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 $z=5i$:


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


as was required.


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

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


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

  • $\begingroup$ Wait so how you integrate that. And how do you take the limit? $\endgroup$
    – Ivan
    Oct 30, 2014 at 1:39
  • $\begingroup$ You find an antiderivative, see the other answers for that. Then you take the limit. $\endgroup$
    – Joao
    Oct 30, 2014 at 1:46

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

  • $\begingroup$ Hmm really not sure how to deal with those integrals neither. Are you doing the same method as mookid? Also how can the parameter be x when value is z^2. $\endgroup$
    – Ivan
    Oct 30, 2014 at 2:00
  • $\begingroup$ I think mookid is deriving the formula I was referring to. When you're dealing directly with Trig Integrals you might have to show the derivation. However, when dealing with Indefinite Integrals most people assume you know how to derive the answer, and thus let you use the formula. Similar to how in Calc 1 you had to use the formal definition of a derivative before being able to use the tricks. $\endgroup$
    – steveclark
    Oct 30, 2014 at 2:08
  • $\begingroup$ As for the difference between $x$ and $z$. $\,\,x$ is only used in the bounds of the integral, where as $z$ is bounded to the function that is being integrated. They are separate. $\endgroup$
    – steveclark
    Oct 30, 2014 at 2:10

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 $$\lim_{u \to \pm \infty} \arctan u = \pm \frac{\pi}{2} .$$

  • $\begingroup$ how did you ge the parameter -pi/2 to pi/2 $\endgroup$
    – Ivan
    Oct 30, 2014 at 1:49
  • $\begingroup$ Simply use the limit definition of an integral with infinite limit(s) and use the limit in the answer. $\endgroup$ Oct 30, 2014 at 1:55

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

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

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


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


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