Calculate the improper integral: integral from 1 to ()

Calculate the improper integral: $$ \int_1^\infty \frac{dt}{t\sqrt{t^2+a^2}} \quad (a \neq 0) $$

This is an improper integral at infinity with a parameter $a$.

Step 1: Express as a limit

$$\int_1^\infty \frac{dt}{t\sqrt{t^2+a^2}} = \lim_{R \to \infty} \int_1^R \frac{dt}{t\sqrt{t^2+a^2}}$$

Step 2: Find the antiderivative using substitution

For the integral $\int \frac{dt}{t\sqrt{t^2+a^2}}$, let's use the substitution:

Let $u = \sqrt{t^2 + a^2}$, so $u^2 = t^2 + a^2$, which gives $t^2 = u^2 - a^2$.

Then: $2u \, du = 2t \, dt$, so $t \, dt = u \, du$

Also: $dt = \frac{u \, du}{t} = \frac{u \, du}{\sqrt{u^2 - a^2}}$

The integral becomes:

$$\int \frac{dt}{t\sqrt{t^2+a^2}} = \int \frac{1}{\sqrt{u^2-a^2} \cdot u} \cdot \frac{u \, du}{\sqrt{u^2-a^2}}$$

$$= \int \frac{du}{u^2 - a^2}$$

Step 3: Evaluate using partial fractions

$$\frac{1}{u^2 - a^2} = \frac{1}{(u-a)(u+a)} = \frac{A}{u-a} + \frac{B}{u+a}$$

Multiplying by $(u-a)(u+a)$:

$$1 = A(u+a) + B(u-a)$$

Setting $u = a$: $1 = 2aA \implies A = \frac{1}{2a}$

Setting $u = -a$: $1 = -2aB \implies B = -\frac{1}{2a}$

Therefore:

$$\frac{1}{u^2 - a^2} = \frac{1}{2a}\left(\frac{1}{u-a} - \frac{1}{u+a}\right)$$

So:

$$\int \frac{du}{u^2 - a^2} = \frac{1}{2a}\left[\ln|u-a| - \ln|u+a|\right] + C$$

$$= \frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right| + C$$

Substituting back $u = \sqrt{t^2 + a^2}$:

$$= \frac{1}{2a}\ln\left|\frac{\sqrt{t^2+a^2} - a}{\sqrt{t^2+a^2} + a}\right| + C$$

Step 4: Simplify using rationalization

We can also write this using the formula:

$$\int \frac{dt}{t\sqrt{t^2+a^2}} = \frac{1}{|a|}\ln\left|\frac{\sqrt{t^2+a^2} - |a|}{t}\right| + C$$

For convenience with $a \neq 0$, let's work with $|a|$ explicitly.

Actually, let's use a different form. The standard result is:

$$\int \frac{dt}{t\sqrt{t^2+a^2}} = -\frac{1}{|a|}\text{arcsec}\left(\frac{|t|}{|a|}\right) + C$$

or equivalently:

$$= \frac{1}{|a|}\ln\left|\frac{|a|}{t + \sqrt{t^2+a^2}}\right| + C$$

Let's use the logarithmic form which is more straightforward.

Step 5: Alternative - Direct trigonometric substitution

Let $t = |a|\sec\theta$, so $dt = |a|\sec\theta\tan\theta \, d\theta$

Then: $\sqrt{t^2 + a^2} = \sqrt{a^2\sec^2\theta + a^2} = |a|\sqrt{\sec^2\theta + 1} = |a|\tan\theta$ (for appropriate range)

Actually, this becomes messy. Let's stick with the result from Step 3.

Step 6: Evaluate the definite integral

Using the antiderivative from Step 3:

$$\int_1^R \frac{dt}{t\sqrt{t^2+a^2}} = \left[\frac{1}{2|a|}\ln\left|\frac{\sqrt{t^2+a^2} - |a|}{\sqrt{t^2+a^2} + |a|}\right|\right]_1^R$$

Wait, let me reconsider. For $t > 0$ and using $a^2 > 0$:

$$\int \frac{dt}{t\sqrt{t^2+a^2}} = \frac{1}{|a|}\ln\left(\frac{t}{|a| + \sqrt{t^2+a^2}}\right) + C$$

Actually, the cleanest form is:

$$\int \frac{dt}{t\sqrt{t^2+a^2}} = -\frac{1}{|a|}\text{arcsinh}\left(\frac{|a|}{t}\right) + C$$

Let me use the most standard form:

$$\int \frac{dt}{t\sqrt{t^2+a^2}} = \frac{1}{|a|}\ln\left|\frac{t}{|a| + \sqrt{t^2+a^2}}\right| + C$$

Step 7: Evaluate at the limits

At $t = R$:

$$\frac{1}{|a|}\ln\left(\frac{R}{|a| + \sqrt{R^2+a^2}}\right)$$

At $t = 1$:

$$\frac{1}{|a|}\ln\left(\frac{1}{|a| + \sqrt{1+a^2}}\right)$$

Therefore:

$$\int_1^R \frac{dt}{t\sqrt{t^2+a^2}} = \frac{1}{|a|}\ln\left(\frac{R}{|a| + \sqrt{R^2+a^2}}\right) - \frac{1}{|a|}\ln\left(\frac{1}{|a| + \sqrt{1+a^2}}\right)$$

$$= \frac{1}{|a|}\ln\left(\frac{R(|a| + \sqrt{1+a^2})}{|a| + \sqrt{R^2+a^2}}\right)$$

Step 8: Take the limit as $R \to \infty$

$$\lim_{R \to \infty} \frac{1}{|a|}\ln\left(\frac{R(|a| + \sqrt{1+a^2})}{|a| + \sqrt{R^2+a^2}}\right)$$

For large $R$: $\sqrt{R^2+a^2} \approx R\sqrt{1 + a^2/R^2} \approx R$

More precisely:

$$\sqrt{R^2+a^2} = R\sqrt{1 + a^2/R^2} = R\left(1 + \frac{a^2}{2R^2} + O(1/R^4)\right)$$

Therefore:

$$\frac{R}{|a| + \sqrt{R^2+a^2}} = \frac{R}{|a| + R(1 + \frac{a^2}{2R^2} + ...)} \approx \frac{R}{|a| + R} \to 1$$

More carefully:

$$\frac{R}{|a| + \sqrt{R^2+a^2}} = \frac{1}{\frac{|a|}{R} + \sqrt{1 + \frac{a^2}{R^2}}} \to \frac{1}{0 + 1} = 1$$

Therefore:

$$\lim_{R \to \infty} \ln\left(\frac{R(|a| + \sqrt{1+a^2})}{|a| + \sqrt{R^2+a^2}}\right) = \ln(|a| + \sqrt{1+a^2})$$

Final Answer

$$\boxed{\int_1^\infty \frac{dt}{t\sqrt{t^2+a^2}} = \frac{1}{|a|}\ln(|a| + \sqrt{1+a^2})}$$

The integral converges for all $a \neq 0$.

Alternative form using $\text{arcsinh}$

Since $\text{arcsinh}(x) = \ln(x + \sqrt{1+x^2})$, we can write:

$$\int_1^\infty \frac{dt}{t\sqrt{t^2+a^2}} = \frac{1}{|a|}\text{arcsinh}\left(\frac{|a|}{1}\right) = \frac{1}{|a|}\text{arcsinh}(|a|)$$

Special cases

Case 1: $a = 1$

$$\int_1^\infty \frac{dt}{t\sqrt{t^2+1}} = \ln(1 + \sqrt{2}) = \text{arcsinh}(1) \approx 0.881$$

Case 2: $a = -1$ (same as $a = 1$ since we use $|a|$)

$$\int_1^\infty \frac{dt}{t\sqrt{t^2+1}} = \ln(1 + \sqrt{2}) \approx 0.881$$

Case 3: $a = 2$

$$\int_1^\infty \frac{dt}{t\sqrt{t^2+4}} = \frac{1}{2}\ln(2 + \sqrt{5}) \approx 0.440$$

Why the integral converges

For large $t$:

$$\frac{1}{t\sqrt{t^2+a^2}} \approx \frac{1}{t \cdot t} = \frac{1}{t^2}$$

Since $\int_1^\infty \frac{dt}{t^2}$ converges (exponent 2 > 1), our integral also converges. ✓

Behavior as $|a|$ varies

The integral value $\frac{1}{|a|}\ln(|a| + \sqrt{1+a^2})$:

• As $|a| \to 0^+$: $\ln(|a| + \sqrt{1+a^2}) \approx \ln(|a| + 1) \to \ln 1 = 0$, but divided by $|a| \to 0$, so we need L'Hôpital's rule. Actually, for small $|a|$: $\ln(|a| + \sqrt{1+a^2}) \approx \ln(1 + |a|) \approx |a|$ So: $\frac{|a|}{|a|} \to 1$ as $|a| \to 0$
• As $|a| \to \infty$: $\ln(|a| + \sqrt{1+a^2}) \approx \ln(2|a|)$ So: $\frac{\ln(2|a|)}{|a|} \to 0$

Summary

1. The integral is improper at $t = \infty$
2. Substitution $u = \sqrt{t^2+a^2}$ leads to partial fractions
3. The antiderivative involves logarithms or inverse hyperbolic functions
4. Evaluating limits gives: $\frac{1}{|a|}\ln(|a| + \sqrt{1+a^2})$
5. Equivalently: $\frac{1}{|a|}\text{arcsinh}(|a|)$
6. The integral converges because of $\frac{1}{t^2}$ decay at infinity

Key lessons:

• Integrals of the form $\frac{1}{t\sqrt{t^2+a^2}}$ arise in many applications
• The result involves inverse hyperbolic sine (arcsinh)
• The answer depends on $|a|$, not the sign of $a$
• Convergence assured by $\frac{1}{t^2}$ behavior at infinity
• Beautiful connection to hyperbolic functions