Question

Find the leading behavior as x→0+ of the following integrals:

(a) ∫ ₓ¹ cos(xt)dt
(b) ∫₀¹​ √sinh(xt)​ dt;
(c) ∫₀¹​ e −ˣ/ᵗ dt;
(d) ∫ₓ¹ e −ˣ/ᵗ dt
(e) ∫ ₓ¹ sin(xt)dt
(f) ∫₀¹ [dt/(1−t)](eˣ −eˣᵗ );
(g) ∫₀¹ /x e −ᵗ² dt;
(h) ∫₀¹ [e −ˣᵗ /(1+t² )]dt;
(i) ∫₀[infinity] J ₀(xt)t −¹/² dt.

Answer 1

The leading behavior as x approaches 0+ of the given integrals can be determined by analyzing the exponent of the variable x in the integrand.

Step-by-step explanation:

The leading behavior of an integral as x approaches 0+ can be determined by analyzing the exponent of the variable x in the integrand:

(a) ∫ₓ¹ cos(xt)dt:

As x approaches 0+, the integral ∫ₓ¹ cos(xt)dt approaches 1 since the cosine function oscillates between -1 and 1.

(b) ∫₀¹​ √sinh(xt)​ dt:

As x approaches 0+, the hyperbolic sine function sinh(xt) approaches 0 and the integral ∫₀¹​ √sinh(xt)​ dt also approaches 0.

(c) ∫₀¹​ e^-ᵗ/x dt:

As x approaches 0+, the exponential term e^-ᵗ/x approaches 1 and the integral ∫₀¹​ e^-ᵗ/x dt approaches 1.

(d) ∫ₓ¹ e^-ᵗ/x dt:

As x approaches 0+, the exponential term e^-ᵗ/x approaches infinity and the integral ∫ₓ¹ e^-ᵗ/x dt approaches infinity.

(e) ∫ₓ¹ sin(xt)dt:

As x approaches 0+, the integral ∫ₓ¹ sin(xt)dt approaches 0 since the sine function oscillates around 0.

(f) ∫₀¹ [dt/(1-t)](e^x - e^xt):

As x approaches 0+, the integral ∫₀¹ [dt/(1-t)](e^x - e^xt) approaches -∞.

(g) ∫₀¹ /x e^(-t²) dt:

As x approaches 0+, the integral ∫₀¹ /x e^(-t²) dt approaches ∞.

(h) ∫₀¹ [e^-xt /(1+t²)]dt:

As x approaches 0+, the integral ∫₀¹ [e^-xt /(1+t²)]dt approaches π/2.

(i) ∫₀ to infinity J₀(xt)t^(-1/2) dt:

As x approaches 0+, the integral ∫₀ to infinity J₀(xt)t^(-1/2) dt approaches 0.